Vector Algebra
273 Questions
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th June Evening Shift
Let $\overrightarrow b = \widehat i + \widehat j + \lambda \widehat k$, $\lambda$ $\in$ R. If $\overrightarrow a $ is a vector such that $\overrightarrow a \times \overrightarrow b = 13\widehat i - \widehat j - 4\widehat k$ and $\overrightarrow a \,.\,\overrightarrow b + 21 = 0$, then $\left( {\overrightarrow b - \overrightarrow a } \right).\,\left( {\widehat k - \widehat j} \right) + \left( {\overrightarrow b + \overrightarrow a } \right).\,\left( {\widehat i - \widehat k} \right)$ is equal to _____________.
Correct Answer: 14
Explanation:
Let $\overrightarrow a = x\widehat i = y\widehat j + z\widehat k$
So, $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr x & y & z \cr 1 & 1 & \lambda \cr } } \right| = \widehat i(\lambda y - z) + \widehat j(z - \lambda x) + \widehat k(x - y)$
$ \Rightarrow \lambda y - z = 13,\,z - \lambda x = - 1,\,x - y = - 4$
and $x + y + \lambda z = - 21$
$\Rightarrow$ Clearly, $\lambda = 3$, $x = - 2$, $y = 2$ and $z = - 7$
So, $\overrightarrow b - \overrightarrow a = 3\widehat i - \widehat j + 10\widehat k$
and $\overrightarrow b + \overrightarrow a = - \widehat i + 3\widehat j - 4\widehat k$
$ \Rightarrow \left( {\overrightarrow b - \overrightarrow a } \right)\,.\,\left( {\widehat k - \widehat j} \right) + \left( {\overrightarrow b + \overrightarrow a } \right)\,.\,\left( {\widehat i - \widehat k} \right) = 11 + 3 = 14$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th June Morning Shift
Let $\theta$ be the angle between the vectors $\overrightarrow a $ and $\overrightarrow b $, where $|\overrightarrow a | = 4,$ $|\overrightarrow b | = 3$ and $\theta \in \left( {{\pi \over 4},{\pi \over 3}} \right)$. Then ${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$ is equal to __________.
Correct Answer: 576
Explanation:
${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$
$ \Rightarrow {\left| {\overrightarrow a \times \overrightarrow a + \overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow a - \overrightarrow b \times \overrightarrow b } \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$
$ \Rightarrow {\left| {2\left( {\overrightarrow a \times \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$
$ \Rightarrow 4{\left( {\overrightarrow a \times \overrightarrow b } \right)^2} + {\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$
$ \Rightarrow 4{\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2} = 4\,.\,16\,.\,9 = 576$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Evening Shift
Let $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ three vectors mutually perpendicular to each other and have same magnitude. If a vector ${ \overrightarrow r } $ satisfies.
$\overrightarrow a \times \{ (\overrightarrow r - \overrightarrow b ) \times \overrightarrow a \} + \overrightarrow b \times \{ (\overrightarrow r - \overrightarrow c ) \times \overrightarrow b \} + \overrightarrow c \times \{ (\overrightarrow r - \overrightarrow a ) \times \overrightarrow c \} = \overrightarrow 0 $, then $\overrightarrow r $ is equal to :
$\overrightarrow a \times \{ (\overrightarrow r - \overrightarrow b ) \times \overrightarrow a \} + \overrightarrow b \times \{ (\overrightarrow r - \overrightarrow c ) \times \overrightarrow b \} + \overrightarrow c \times \{ (\overrightarrow r - \overrightarrow a ) \times \overrightarrow c \} = \overrightarrow 0 $, then $\overrightarrow r $ is equal to :
A.
${1 \over 3}(\overrightarrow a + \overrightarrow b + \overrightarrow c )$
B.
${1 \over 3}(2\overrightarrow a + \overrightarrow b - \overrightarrow c )$
C.
${1 \over 2}(\overrightarrow a + \overrightarrow b + \overrightarrow c )$
D.
${1 \over 2}(\overrightarrow a + \overrightarrow b + 2\overrightarrow c )$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Morning Shift
Let $\overrightarrow a $ and $\overrightarrow b $ be two vectors
such that $\left| {2\overrightarrow a + 3\overrightarrow b } \right| = \left| {3\overrightarrow a + \overrightarrow b } \right|$ and the angle between $\overrightarrow a $ and $\overrightarrow b $ is 60$^\circ$. If ${1 \over 8}\overrightarrow a $ is a unit vector, then $\left| {\overrightarrow b } \right|$ is equal to :
such that $\left| {2\overrightarrow a + 3\overrightarrow b } \right| = \left| {3\overrightarrow a + \overrightarrow b } \right|$ and the angle between $\overrightarrow a $ and $\overrightarrow b $ is 60$^\circ$. If ${1 \over 8}\overrightarrow a $ is a unit vector, then $\left| {\overrightarrow b } \right|$ is equal to :
A.
4
B.
6
C.
5
D.
8
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Evening Shift
A hall has a square floor of dimension 10 m $\times$ 10 m (see the figure) and vertical walls. If the angle GPH between the diagonals AG and BH is ${\cos ^{ - 1}}{1 \over 5}$, then the height of the hall (in meters) is :
A.
5
B.
2$\sqrt {10} $
C.
5$\sqrt {3} $
D.
5$\sqrt {2} $
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
Let $\overrightarrow a = \widehat i + \widehat j + \widehat k$ and $\overrightarrow b = \widehat j - \widehat k$. If $\overrightarrow c $ is a vector such that $\overrightarrow a \times \overrightarrow c = \overrightarrow b $ and $\overrightarrow a .\overrightarrow c = 3$, then $\overrightarrow a .(\overrightarrow b \times \overrightarrow c )$ is equal to :
A.
$-$2
B.
$-$6
C.
6
D.
2
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Evening Shift
Let $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ be three vectors such that $\overrightarrow a $ = $\overrightarrow b $ $\times$ ($\overrightarrow b $ $\times$ $\overrightarrow c $). If magnitudes of the vectors $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ are $\sqrt 2 $, 1 and 2 respectively and the angle between $\overrightarrow b $ and $\overrightarrow c $ is $\theta \left( {0 < \theta < {\pi \over 2}} \right)$, then the value of 1 + tan$\theta$ is equal to :
A.
$\sqrt 3 + 1$
B.
2
C.
1
D.
${{\sqrt 3 + 1} \over {\sqrt 3 }}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Morning Shift
Let $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$ and $\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$. Then the vector product $\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\left( {\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightarrow b } \right) \times \overrightarrow b } \right)} \right) \times \overrightarrow b } \right)$ is equal to :
A.
$5(34\widehat i - 5\widehat j + 3\widehat k)$
B.
$7(34\widehat i - 5\widehat j + 3\widehat k)$
C.
$7(30\widehat i - 5\widehat j + 7\widehat k)$
D.
$5(30\widehat i - 5\widehat j + 7\widehat k)$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
Let a, b and c be distinct positive numbers. If the vectors $a\widehat i + a\widehat j + c\widehat k,\widehat i+\widehat k$ and $c\widehat i + c\widehat j + b\widehat k$ are co-planar, then c is equal to :
A.
${2 \over {{1 \over a} + {1 \over b}}}$
B.
${{a + b} \over 2}$
C.
${1 \over a} + {1 \over b}$
D.
$\sqrt {ab} $
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
If $\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$ and $\left| {\overrightarrow a \times \overrightarrow b } \right|$ = 8, then $\left| {\overrightarrow a .\,\overrightarrow b } \right|$ is equal to :
A.
6
B.
4
C.
3
D.
5
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Morning Shift
Let the vectors
$(2 + a + b)\widehat i + (a + 2b + c)\widehat j - (b + c)\widehat k,(1 + b)\widehat i + 2b\widehat j - b\widehat k$ and $(2 + b)\widehat i + 2b\widehat j + (1 - b)\widehat k$, $a,b,c, \in R$
be co-planar. Then which of the following is true?
$(2 + a + b)\widehat i + (a + 2b + c)\widehat j - (b + c)\widehat k,(1 + b)\widehat i + 2b\widehat j - b\widehat k$ and $(2 + b)\widehat i + 2b\widehat j + (1 - b)\widehat k$, $a,b,c, \in R$
be co-planar. Then which of the following is true?
A.
2b = a + c
B.
3c = a + b
C.
a = b + 2c
D.
2a = b + c
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
Let a vector ${\overrightarrow a }$ be coplanar with vectors $\overrightarrow b = 2\widehat i + \widehat j + \widehat k$ and $\overrightarrow c = \widehat i - \widehat j + \widehat k$. If ${\overrightarrow a}$ is perpendicular to $\overrightarrow d = 3\widehat i + 2\widehat j + 6\widehat k$, and $\left| {\overrightarrow a } \right| = \sqrt {10} $. Then a possible value of $[\matrix{
{\overrightarrow a } & {\overrightarrow b } & {\overrightarrow c } \cr
} ] + [\matrix{
{\overrightarrow a } & {\overrightarrow b } & {\overrightarrow d } \cr
} ] + [\matrix{
{\overrightarrow a } & {\overrightarrow c } & {\overrightarrow d } \cr
} ]$ is equal to :
A.
$-$42
B.
$-$40
C.
$-$29
D.
$-$38
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
Let three vectors $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ be such that $\overrightarrow a \times \overrightarrow b = \overrightarrow c $, $\overrightarrow b \times \overrightarrow c = \overrightarrow a $ and $\left| {\overrightarrow a } \right| = 2$. Then which one of the following is not true?
A.
$\overrightarrow a \times \left( {(\overrightarrow b + \overrightarrow c ) \times (\overrightarrow b \times \overrightarrow c )} \right) = \overrightarrow 0 $
B.
Projection of $\overrightarrow a $ on $(\overrightarrow b \times \overrightarrow c )$ is 2
C.
$\left[ {\matrix{
{\overrightarrow a } & {\overrightarrow b } & {\overrightarrow c } \cr
} } \right] + \left[ {\matrix{
{\overrightarrow c } & {\overrightarrow a } & {\overrightarrow b } \cr
} } \right] = 8$
D.
${\left| {3\overrightarrow a + \overrightarrow b - 2\overrightarrow c } \right|^2} = 51$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Evening Shift
In a triangle ABC, if $\left| {\overrightarrow {BC} } \right| = 3$, $\left| {\overrightarrow {CA} } \right| = 5$ and $\left| {\overrightarrow {BA} } \right| = 7$, then the projection of the vector $\overrightarrow {BA} $ on $\overrightarrow {BC} $ is equal to :
A.
${{19} \over 2}$
B.
${{13} \over 2}$
C.
${{11} \over 2}$
D.
${{15} \over 2}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Morning Shift
Let $\overrightarrow a = 2\widehat i + \widehat j - 2\widehat k$ and $\overrightarrow b = \widehat i + \widehat j$. If $\overrightarrow c $ is a vector such that $\overrightarrow a .\,\overrightarrow c = \left| {\overrightarrow c } \right|,\left| {\overrightarrow c - \overrightarrow a } \right| = 2\sqrt 2 $ and the angle between $(\overrightarrow a \times \overrightarrow b )$ and $\overrightarrow c $ is ${\pi \over 6}$, then the value of $\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right|$ is :
A.
${2 \over 3}$
B.
4
C.
3
D.
${3 \over 2}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Evening Shift
Let $\overrightarrow a $ and $\overrightarrow b $ be two non-zero vectors perpendicular to each other and $|\overrightarrow a | = |\overrightarrow b |$. If $|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a |$, then the angle between the vectors $\left( {\overrightarrow a + \overrightarrow b + \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)$ and ${\overrightarrow a }$ is equal to :
A.
${\sin ^{ - 1}}\left( {{1 \over {\sqrt 6 }}} \right)$
B.
${\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right)$
C.
${\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$
D.
${\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Evening Shift
In a triangle ABC, if $|\overrightarrow {BC} | = 8,|\overrightarrow {CA} | = 7,|\overrightarrow {AB} | = 10$, then the projection of the vector $\overrightarrow {AB} $ on $\overrightarrow {AC} $ is equal to :
A.
${{25} \over 4}$
B.
${{127} \over 20}$
C.
${{85} \over 14}$
D.
${{115} \over 16}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
A vector $\overrightarrow a $ has components 3p and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, $\overrightarrow a $ has components p + 1 and $\sqrt {10} $, then the value of p is equal to :
A.
1
B.
$ - {5 \over 4}$
C.
${4 \over 5}$
D.
$-$1
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
Let O be the origin. Let $\overrightarrow {OP} = x\widehat i + y\widehat j - \widehat k$ and $\overrightarrow {OQ} = - \widehat i + 2\widehat j + 3x\widehat k$, x, y$\in$R, x > 0, be such that $\left| {\overrightarrow {PQ} } \right| = \sqrt {20} $ and the vector $\overrightarrow {OP} $ is perpendicular $\overrightarrow {OQ} $. If $\overrightarrow {OR} $ = $3\widehat i + z\widehat j - 7\widehat k$, z$\in$R, is coplanar with $\overrightarrow {OP} $ and $\overrightarrow {OQ} $, then the value of x2 + y2 + z2 is equal to :
A.
2
B.
9
C.
7
D.
1
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
Let $\overrightarrow a $ = 2$\widehat i$ $-$ 3$\widehat j$ + 4$\widehat k$ and $\overrightarrow b $ = 7$\widehat i$ + $\widehat j$ $-$ 6$\widehat k$.
If $\overrightarrow r $ $\times$ $\overrightarrow a $ = $\overrightarrow r $ $\times$ $\overrightarrow b $, $\overrightarrow r $ . ($\widehat i$ + 2$\widehat j$ + $\widehat k$) = $-$3, then $\overrightarrow r $ . (2$\widehat i$ $-$ 3$\widehat j$ + $\widehat k$) is equal to :
If $\overrightarrow r $ $\times$ $\overrightarrow a $ = $\overrightarrow r $ $\times$ $\overrightarrow b $, $\overrightarrow r $ . ($\widehat i$ + 2$\widehat j$ + $\widehat k$) = $-$3, then $\overrightarrow r $ . (2$\widehat i$ $-$ 3$\widehat j$ + $\widehat k$) is equal to :
A.
10
B.
8
C.
13
D.
12
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Evening Shift
Let $\overrightarrow a $ = $\widehat i$ + 2$\widehat j$ $-$ 3$\widehat k$ and $\overrightarrow b = 2\widehat i$ $-$ 3$\widehat j$ + 5$\widehat k$. If $\overrightarrow r $ $\times$ $\overrightarrow a $ = $\overrightarrow b $ $\times$ $\overrightarrow r $,
$\overrightarrow r $ . $\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right)$ = 3 and $\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right)$ = $-$1, $\alpha$ $\in$ R, then the
value of $\alpha$ + ${\left| {\overrightarrow r } \right|^2}$ is equal to :
$\overrightarrow r $ . $\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right)$ = 3 and $\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right)$ = $-$1, $\alpha$ $\in$ R, then the
value of $\alpha$ + ${\left| {\overrightarrow r } \right|^2}$ is equal to :
A.
13
B.
11
C.
9
D.
15
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
Let a vector $\alpha \widehat i + \beta \widehat j$ be obtained by rotating the vector $\sqrt 3 \widehat i + \widehat j$ by an angle 45$^\circ$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices ($\alpha$, $\beta$), (0, $\beta$) and (0, 0) is equal to :
A.
${1 \over {\sqrt 2 }}$
B.
${1 \over 2}$
C.
1
D.
2${\sqrt 2 }$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Evening Shift
If vectors $\overrightarrow {{a_1}} = x\widehat i - \widehat j + \widehat k$ and $\overrightarrow {{a_2}} = \widehat i + y\widehat j + z\widehat k$ are collinear, then a possible unit vector parallel to the vector $x\widehat i + y\widehat j + z\widehat k$ is :
A.
${1 \over {\sqrt 3 }}\left( {\widehat i - \widehat j + \widehat k} \right)$
B.
${1 \over {\sqrt 2 }}\left( { - \widehat j + \widehat k} \right)$
C.
${1 \over {\sqrt 2 }}\left( {\widehat i - \widehat j} \right)$
D.
${1 \over {\sqrt 3 }}\left( {\widehat i + \widehat j - \widehat k} \right)$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
If $\overrightarrow a $ and $\overrightarrow b $ are perpendicular, then
$\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)} \right)$ is equal to :
$\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)} \right)$ is equal to :
A.
${1 \over 2}|\overrightarrow a {|^4}\overrightarrow b $
B.
$\overrightarrow 0 $
C.
$\overrightarrow a \times \overrightarrow b $
D.
$|\overrightarrow a {|^4}\overrightarrow b $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
Let $\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k$ and $\overrightarrow b = \widehat i + 2\widehat j - \widehat k$. Let a vector $\overrightarrow v $ be in the plane containing $\overrightarrow a $ and $\overrightarrow b $. If $\overrightarrow v $ is perpendicular to the vector $3\widehat i + 2\widehat j - \widehat k$ and its projection on $\overrightarrow a $ is 19 units, then ${\left| {2\overrightarrow v } \right|^2}$ is equal to _____________.
Correct Answer: 1494
Explanation:
$\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k$
$\overrightarrow b = \widehat i + 2\widehat j - \widehat k$
$\overrightarrow c = 3\widehat i + 2\widehat j - \widehat k$
$\overrightarrow v = x\overrightarrow a + y\overrightarrow b $
$\overrightarrow v \left( {3\widehat i + 2\widehat j - \widehat k} \right) = 0$
$\overrightarrow v .\widehat a = 19$
$\overrightarrow v = \lambda \overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right)$
$\overrightarrow v = \lambda \left[ {\left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b } \right]$
$ = \lambda [(3 + 4 + 1)\left( {2\widehat i - \widehat j + 2\widehat k} \right) - \left( {{{6 - 2 - 2} \over 2}} \right)\left( {\widehat i + 2\widehat j + \widehat k} \right)$
$ = \lambda [16\widehat i - 8\widehat j + 16\widehat k - 2\widehat i - 4\widehat j + 2\widehat k]$
$\overrightarrow v = \lambda \left[ {14\widehat i - 12\widehat j + 18\widehat k} \right]$
$\lambda [14\widehat i - 12\widehat j + 18\widehat k].{{\left( {2\widehat i - \widehat j + 2\widehat k} \right)} \over {\sqrt {4 + 1 + 4} }} = 19$
$\lambda {{[28 + 12 + 36]} \over 3} = 19$
$\lambda \left( {{{76} \over 3}} \right) = 19$
$4\lambda = 3 \Rightarrow \lambda = {3 \over 4}$
$|2{v^2}| = {\left| {2 \times {3 \over 4}(14\widehat i - 12\widehat j + 18\widehat k)} \right|^2}$
${9 \over 4} \times 4{\left( {7\widehat i - 6\widehat j + 9\widehat k} \right)^2}$
$ = 9(49 + 36 + 81)$
$ = 9(166)$
$ = 1494$
$\overrightarrow b = \widehat i + 2\widehat j - \widehat k$
$\overrightarrow c = 3\widehat i + 2\widehat j - \widehat k$
$\overrightarrow v = x\overrightarrow a + y\overrightarrow b $
$\overrightarrow v \left( {3\widehat i + 2\widehat j - \widehat k} \right) = 0$
$\overrightarrow v .\widehat a = 19$
$\overrightarrow v = \lambda \overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right)$
$\overrightarrow v = \lambda \left[ {\left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b } \right]$
$ = \lambda [(3 + 4 + 1)\left( {2\widehat i - \widehat j + 2\widehat k} \right) - \left( {{{6 - 2 - 2} \over 2}} \right)\left( {\widehat i + 2\widehat j + \widehat k} \right)$
$ = \lambda [16\widehat i - 8\widehat j + 16\widehat k - 2\widehat i - 4\widehat j + 2\widehat k]$
$\overrightarrow v = \lambda \left[ {14\widehat i - 12\widehat j + 18\widehat k} \right]$
$\lambda [14\widehat i - 12\widehat j + 18\widehat k].{{\left( {2\widehat i - \widehat j + 2\widehat k} \right)} \over {\sqrt {4 + 1 + 4} }} = 19$
$\lambda {{[28 + 12 + 36]} \over 3} = 19$
$\lambda \left( {{{76} \over 3}} \right) = 19$
$4\lambda = 3 \Rightarrow \lambda = {3 \over 4}$
$|2{v^2}| = {\left| {2 \times {3 \over 4}(14\widehat i - 12\widehat j + 18\widehat k)} \right|^2}$
${9 \over 4} \times 4{\left( {7\widehat i - 6\widehat j + 9\widehat k} \right)^2}$
$ = 9(49 + 36 + 81)$
$ = 9(166)$
$ = 1494$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
Let $\overrightarrow a = \widehat i + 5\widehat j + \alpha \widehat k$, $\overrightarrow b = \widehat i + 3\widehat j + \beta \widehat k$ and $\overrightarrow c = - \widehat i + 2\widehat j - 3\widehat k$ be three vectors such that, $\left| {\overrightarrow b \times \overrightarrow c } \right| = 5\sqrt 3 $ and ${\overrightarrow a }$ is perpendicular to ${\overrightarrow b }$. Then the greatest amongst the values of ${\left| {\overrightarrow a } \right|^2}$ is _____________.
Correct Answer: 90
Explanation:
Since, $\overrightarrow a .\,\overrightarrow b = 0$
$1 + 15 + \alpha \beta = 0 \Rightarrow \alpha \beta = - 16$ .... (1)
Also,
${\left| {\overrightarrow b \, \times \overrightarrow c } \right|^2} = 75 \Rightarrow (10 + {\beta ^2})14 - {(5 - 3\beta )^2} = 75$
$\Rightarrow$ 5$\beta$2 + 30$\beta$ + 40 = 0
$\Rightarrow$ $\beta$ = $-$4, $-$2
$\Rightarrow$ $\alpha$ = 4, 8
$ \Rightarrow \left| {\overrightarrow a } \right|_{\max }^2 = {(26 + {\alpha ^2})_{\max }} = 90$
$1 + 15 + \alpha \beta = 0 \Rightarrow \alpha \beta = - 16$ .... (1)
Also,
${\left| {\overrightarrow b \, \times \overrightarrow c } \right|^2} = 75 \Rightarrow (10 + {\beta ^2})14 - {(5 - 3\beta )^2} = 75$
$\Rightarrow$ 5$\beta$2 + 30$\beta$ + 40 = 0
$\Rightarrow$ $\beta$ = $-$4, $-$2
$\Rightarrow$ $\alpha$ = 4, 8
$ \Rightarrow \left| {\overrightarrow a } \right|_{\max }^2 = {(26 + {\alpha ^2})_{\max }} = 90$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
If the projection of the vector $\widehat i + 2\widehat j + \widehat k$ on the sum of the two vectors $2\widehat i + 4\widehat j - 5\widehat k$ and $ - \lambda \widehat i + 2\widehat j + 3\widehat k$ is 1, then $\lambda$ is equal to __________.
Correct Answer: 5
Explanation:
$\overrightarrow a = \widehat i + 2\widehat j + \widehat k$
$\overrightarrow b = (2 - \lambda )\widehat i + 6\widehat j - 2\widehat k$
${{\overrightarrow a \,.\,\overrightarrow b } \over {|\overrightarrow b |}} = 1,\overrightarrow a \,.\,\overrightarrow b = 12 - \lambda $
$\left( {\overrightarrow a \,.\,\overrightarrow b } \right) = |\overrightarrow b {|^2}$
$\lambda$2 $-$ 24$\lambda$ + 144 = $\lambda$2 $-$ 4$\lambda$ + 4 + 40
20$\lambda$ = 100 $\Rightarrow$ $\lambda$ = 5
$\overrightarrow b = (2 - \lambda )\widehat i + 6\widehat j - 2\widehat k$
${{\overrightarrow a \,.\,\overrightarrow b } \over {|\overrightarrow b |}} = 1,\overrightarrow a \,.\,\overrightarrow b = 12 - \lambda $
$\left( {\overrightarrow a \,.\,\overrightarrow b } \right) = |\overrightarrow b {|^2}$
$\lambda$2 $-$ 24$\lambda$ + 144 = $\lambda$2 $-$ 4$\lambda$ + 4 + 40
20$\lambda$ = 100 $\Rightarrow$ $\lambda$ = 5
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
Let $\overrightarrow a = \widehat i - \alpha \widehat j + \beta \widehat k$, $\overrightarrow b = 3\widehat i + \beta \widehat j - \alpha \widehat k$ and $\overrightarrow c = -\alpha \widehat i - 2\widehat j + \widehat k$, where $\alpha$ and $\beta$ are integers. If $\overrightarrow a \,.\,\overrightarrow b = - 1$ and $\overrightarrow b \,.\,\overrightarrow c = 10$, then $\left( {\overrightarrow a \, \times \overrightarrow b } \right).\,\overrightarrow c $ is equal to ___________.
Correct Answer: 9
Explanation:
$\overrightarrow a = (1, - \alpha ,\beta )$
$\overrightarrow b = (3,\beta , - \alpha )$
$\overrightarrow c = ( - \alpha , - 2,1);\alpha ,\beta \in I$
$\overrightarrow a \,.\,\overrightarrow b = - 1 \Rightarrow 3 - \alpha \beta - \alpha \beta = - 1$
$ \Rightarrow \alpha \beta = 2$
Possible value of
$\alpha $ and $\beta $
$\matrix{ 1 & 2 \cr 2 & 1 \cr { - 1} & { - 2} \cr { - 2} & { - 1} \cr } $
$\overrightarrow b \,.\,\overrightarrow c = 10$
$ \Rightarrow - 3\alpha - 2\beta - \alpha = 10$
$ \Rightarrow 2\alpha + \beta + 5 = 0$
$\therefore$ $\alpha$ = $-$2; $\beta$ = $-$1
$[\overrightarrow a \,\overrightarrow b \,\overrightarrow c ] = \left| {\matrix{ 1 & 2 & { - 1} \cr 3 & { - 1} & 2 \cr 2 & { - 2} & 1 \cr } } \right|$
$ = 1( - 1 + 4) - 2(3 - 4) - 1( - 6 + 2)$
$ = 3 + 2 + 4 = 9$
$\overrightarrow b = (3,\beta , - \alpha )$
$\overrightarrow c = ( - \alpha , - 2,1);\alpha ,\beta \in I$
$\overrightarrow a \,.\,\overrightarrow b = - 1 \Rightarrow 3 - \alpha \beta - \alpha \beta = - 1$
$ \Rightarrow \alpha \beta = 2$
Possible value of
$\alpha $ and $\beta $
$\matrix{ 1 & 2 \cr 2 & 1 \cr { - 1} & { - 2} \cr { - 2} & { - 1} \cr } $
$\overrightarrow b \,.\,\overrightarrow c = 10$
$ \Rightarrow - 3\alpha - 2\beta - \alpha = 10$
$ \Rightarrow 2\alpha + \beta + 5 = 0$
$\therefore$ $\alpha$ = $-$2; $\beta$ = $-$1
$[\overrightarrow a \,\overrightarrow b \,\overrightarrow c ] = \left| {\matrix{ 1 & 2 & { - 1} \cr 3 & { - 1} & 2 \cr 2 & { - 2} & 1 \cr } } \right|$
$ = 1( - 1 + 4) - 2(3 - 4) - 1( - 6 + 2)$
$ = 3 + 2 + 4 = 9$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
Let $\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b $ and $\overrightarrow c = \widehat j - \widehat k$ be three vectors such that $\overrightarrow a \times \overrightarrow b = \overrightarrow c $ and $\overrightarrow a \,.\,\overrightarrow b = 1$. If the length of projection vector of the vector $\overrightarrow b $ on the vector $\overrightarrow a \times \overrightarrow c $ is l, then the value of 3l2 is equal to _____________.
Correct Answer: 2
Explanation:
$\overrightarrow a \times \overrightarrow b = \overrightarrow c $
Take Dot with $\overrightarrow c $
$\left( {\overrightarrow a \times \overrightarrow b } \right).\,\overrightarrow c = {\left| {\overrightarrow c } \right|^2} = 2$
Projection of $\overrightarrow b $ or $\overrightarrow a \times \overrightarrow c = l$
${{\left| {\overrightarrow b \,.\,(\overrightarrow a \times \overrightarrow c )} \right|} \over {|\overrightarrow a \times \overrightarrow c |}} = l$
$\therefore$ $l = {2 \over {\sqrt 6 }} \Rightarrow {l^2} = {4 \over 6}$
$3{l^2} = 2$
Take Dot with $\overrightarrow c $
$\left( {\overrightarrow a \times \overrightarrow b } \right).\,\overrightarrow c = {\left| {\overrightarrow c } \right|^2} = 2$
Projection of $\overrightarrow b $ or $\overrightarrow a \times \overrightarrow c = l$
${{\left| {\overrightarrow b \,.\,(\overrightarrow a \times \overrightarrow c )} \right|} \over {|\overrightarrow a \times \overrightarrow c |}} = l$
$\therefore$ $l = {2 \over {\sqrt 6 }} \Rightarrow {l^2} = {4 \over 6}$
$3{l^2} = 2$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
If $\left( {\overrightarrow a + 3\overrightarrow b } \right)$ is perpendicular to $\left( {7\overrightarrow a - 5\overrightarrow b } \right)$ and $\left( {\overrightarrow a - 4\overrightarrow b } \right)$ is perpendicular to $\left( {7\overrightarrow a - 2\overrightarrow b } \right)$, then the angle between $\overrightarrow a $ and $\overrightarrow b $ (in degrees) is _______________.
Correct Answer: 60
Explanation:
$\left( {\overrightarrow a + 3\overrightarrow b } \right) \bot \left( {7\overrightarrow a - 5\overrightarrow b } \right)$
$ \therefore $ $\left( {\overrightarrow a + 3\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 5\overrightarrow b } \right) = 0$
$ \Rightarrow $ $7{\left| {\overrightarrow a } \right|^2} - 15{\left| {\overrightarrow b } \right|^2} + 16\overrightarrow a \,.\,\overrightarrow b = 0$ ....(1)
Also, $\left( {\overrightarrow a - 4\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 2\overrightarrow b } \right) = 0$
$ \Rightarrow $ $7{\left| {\overrightarrow a } \right|^2} + 8{\left| {\overrightarrow b } \right|^2} - 30\overrightarrow a \,.\,\overrightarrow b = 0$ .....(2)
Equation (1) × 30
$210{\left| {\overrightarrow a } \right|^2} - 450{\left| {\overrightarrow b } \right|^2} + 480\overrightarrow a \,.\,\overrightarrow b = 0$ ....(3)
Equation (2) × 16
$112{\left| {\overrightarrow a } \right|^2} + 128{\left| {\overrightarrow b } \right|^2} - 480\overrightarrow a \,.\,\overrightarrow b = 0$ .....(4)
from (3) & (4)
$322{\left| {\overrightarrow a } \right|^2} = 322{\left| {\overrightarrow b } \right|^2}$
$ \Rightarrow $ ${\left| {\overrightarrow a } \right|^2} = {\left| {\overrightarrow b } \right|^2}$
$ \Rightarrow $ $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|$
From equation (2),
$15\left| {\overrightarrow a } \right| = 30\overrightarrow a .\overrightarrow b $
$ \Rightarrow $ $15{\left| {\overrightarrow a } \right|^2} = 30\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\cos \theta $
$\cos \theta = {{15} \over {30}} = {1 \over 2}$
$\therefore$ $\theta = 60^\circ $
$ \therefore $ $\left( {\overrightarrow a + 3\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 5\overrightarrow b } \right) = 0$
$ \Rightarrow $ $7{\left| {\overrightarrow a } \right|^2} - 15{\left| {\overrightarrow b } \right|^2} + 16\overrightarrow a \,.\,\overrightarrow b = 0$ ....(1)
Also, $\left( {\overrightarrow a - 4\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 2\overrightarrow b } \right) = 0$
$ \Rightarrow $ $7{\left| {\overrightarrow a } \right|^2} + 8{\left| {\overrightarrow b } \right|^2} - 30\overrightarrow a \,.\,\overrightarrow b = 0$ .....(2)
Equation (1) × 30
$210{\left| {\overrightarrow a } \right|^2} - 450{\left| {\overrightarrow b } \right|^2} + 480\overrightarrow a \,.\,\overrightarrow b = 0$ ....(3)
Equation (2) × 16
$112{\left| {\overrightarrow a } \right|^2} + 128{\left| {\overrightarrow b } \right|^2} - 480\overrightarrow a \,.\,\overrightarrow b = 0$ .....(4)
from (3) & (4)
$322{\left| {\overrightarrow a } \right|^2} = 322{\left| {\overrightarrow b } \right|^2}$
$ \Rightarrow $ ${\left| {\overrightarrow a } \right|^2} = {\left| {\overrightarrow b } \right|^2}$
$ \Rightarrow $ $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|$
From equation (2),
$15\left| {\overrightarrow a } \right| = 30\overrightarrow a .\overrightarrow b $
$ \Rightarrow $ $15{\left| {\overrightarrow a } \right|^2} = 30\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\cos \theta $
$\cos \theta = {{15} \over {30}} = {1 \over 2}$
$\therefore$ $\theta = 60^\circ $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
Let $\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$ and $\overrightarrow q = \widehat i + 2\widehat j + \widehat k$ be two vectors. If a vector $\overrightarrow r = (\alpha \widehat i + \beta \widehat j + \gamma \widehat k)$ is perpendicular to each of the vectors ($(\overrightarrow p + \overrightarrow q )$ and $(\overrightarrow p - \overrightarrow q )$, and $\left| {\overrightarrow r } \right| = \sqrt 3 $, then $\left| \alpha \right| + \left| \beta \right| + \left| \gamma \right|$ is equal to _______________.
Correct Answer: 3
Explanation:
$\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$ (Given )
$\overrightarrow q = \widehat i + 2\widehat j + \widehat k$
Now, $(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q ) = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 5 & 2 \cr 1 & 1 & 0 \cr } } \right|$
$ = - 2\widehat i - 2\widehat j - 2\widehat k$
$ \Rightarrow \overrightarrow r = \pm \sqrt 3 {{\left( {(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q )} \right)} \over {\left| {(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q )} \right|}} = \pm {{\sqrt 3 \left( { - 2\widehat i - 2\widehat j - 2\widehat k} \right)} \over {\sqrt {{2^2} + {2^2} + {2^2}} }}$
$\overrightarrow r = \pm \left( { - \widehat i - \widehat j - \widehat k} \right)$
According to question
$\overrightarrow r = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$
So, |$\alpha$| = 1, |$\beta$| = 1, |$\gamma$| = 1
$\Rightarrow$ $\left| \alpha \right| + \left| \beta \right| + \left| \gamma \right|$ = 3
$\overrightarrow q = \widehat i + 2\widehat j + \widehat k$
Now, $(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q ) = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 5 & 2 \cr 1 & 1 & 0 \cr } } \right|$
$ = - 2\widehat i - 2\widehat j - 2\widehat k$
$ \Rightarrow \overrightarrow r = \pm \sqrt 3 {{\left( {(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q )} \right)} \over {\left| {(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q )} \right|}} = \pm {{\sqrt 3 \left( { - 2\widehat i - 2\widehat j - 2\widehat k} \right)} \over {\sqrt {{2^2} + {2^2} + {2^2}} }}$
$\overrightarrow r = \pm \left( { - \widehat i - \widehat j - \widehat k} \right)$
According to question
$\overrightarrow r = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$
So, |$\alpha$| = 1, |$\beta$| = 1, |$\gamma$| = 1
$\Rightarrow$ $\left| \alpha \right| + \left| \beta \right| + \left| \gamma \right|$ = 3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
For p > 0, a vector ${\overrightarrow v _2} = 2\widehat i + (p + 1)\widehat j$ is obtained by rotating the vector ${\overrightarrow v _1} = \sqrt 3 p\widehat i + \widehat j$ by an angle $\theta$ about origin in counter clockwise direction. If $\tan \theta = {{\left( {\alpha \sqrt 3 - 2} \right)} \over {\left( {4\sqrt 3 + 3} \right)}}$, then the value of $\alpha$ is equal to _____________.
Correct Answer: 6
Explanation:
$\left| {\overrightarrow {{V_1}} } \right| = \left| {\overrightarrow {{V_2}} } \right|$
$3{P^2} + 1 = 4 + {(P + 1)^2}$
$2{P^2} - 2P - 4 = 0 \Rightarrow {P^2} - P - 2 = 0$
$P = 2, - 1$ (rejected)
$\cos \theta = {{\overrightarrow {{V_1}} .\overrightarrow {{V_2}} } \over {\left| {\overrightarrow {{V_1}} } \right|\left| {\overrightarrow {{V_2}} } \right|}} = {{2\sqrt 3 P + (P + 1)} \over {\sqrt {{{(P + 1)}^2} + 4} \sqrt {3{P^2} + 1} }}$
$\cos \theta = {{4\sqrt 3 + 3} \over {\sqrt {13} \sqrt {13} }} = {{4\sqrt 3 + 3} \over {13}}$
$\tan \theta = {{\sqrt {112 - 24\sqrt 3 } } \over {4\sqrt 3 + 3}} = {{6\sqrt 3 - 2} \over {4\sqrt 3 + 3}} = {{\alpha \sqrt 3 - 2} \over {4\sqrt 3 + 3}}$
$ \Rightarrow \alpha = 6$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
Let $\overrightarrow a $, $\overrightarrow b $, $\overrightarrow c $ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $\theta$, with the vector $\overrightarrow a $ + $\overrightarrow b $ + $\overrightarrow c $. Then 36cos22$\theta$ is equal to ___________.
Correct Answer: 4
Explanation:
${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2(\overrightarrow a \,.\,\overrightarrow b + \overrightarrow a \,.\,\overrightarrow c + \overrightarrow b \,.\,\overrightarrow c ) = 3$
$ \Rightarrow \left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 3 \overrightarrow a .(\overrightarrow a + \overrightarrow b + \overrightarrow c ) = \left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|\cos \theta $
$ \Rightarrow 1 = \sqrt 3 \cos \theta $
$ \Rightarrow \cos 2\theta = - {1 \over 3}$
$ \Rightarrow 36{\cos ^2}2\theta = 4$
$ \Rightarrow \left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 3 \overrightarrow a .(\overrightarrow a + \overrightarrow b + \overrightarrow c ) = \left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|\cos \theta $
$ \Rightarrow 1 = \sqrt 3 \cos \theta $
$ \Rightarrow \cos 2\theta = - {1 \over 3}$
$ \Rightarrow 36{\cos ^2}2\theta = 4$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
If the shortest distance between the lines $\overrightarrow {{r_1}} = \alpha \widehat i + 2\widehat j + 2\widehat k + \lambda (\widehat i - 2\widehat j + 2\widehat k)$, $\lambda$ $\in$ R, $\alpha$ > 0 and $\overrightarrow {{r_2}} = - 4\widehat i - \widehat k + \mu (3\widehat i - 2\widehat j - 2\widehat k)$, $\mu$ $\in$ R is 9, then $\alpha$ is equal to ____________.
Correct Answer: 6
Explanation:
If $\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b $ and $\overrightarrow r = \overrightarrow c + \lambda \overrightarrow d $ then shortest distance between two lines is
$L = {{(\overrightarrow a - \overrightarrow c ).(\overrightarrow b \times \overrightarrow d )} \over {|b \times d|}}$
$\therefore$ $\overrightarrow a - \overrightarrow c = ((\alpha + 4)\widehat i + 2\widehat j + 3\widehat k)$
${{\overrightarrow b \times \overrightarrow d } \over {|b \times d|}} = {{(2\widehat i + 2\widehat j + \widehat k)} \over 3}$
$\therefore$ $((\alpha + 4)\widehat i + 2\widehat j + 3\widehat k).{{(2\widehat i + 2\widehat j + \widehat k)} \over 3} = 9$
or $\alpha$ = 6
$L = {{(\overrightarrow a - \overrightarrow c ).(\overrightarrow b \times \overrightarrow d )} \over {|b \times d|}}$
$\therefore$ $\overrightarrow a - \overrightarrow c = ((\alpha + 4)\widehat i + 2\widehat j + 3\widehat k)$
${{\overrightarrow b \times \overrightarrow d } \over {|b \times d|}} = {{(2\widehat i + 2\widehat j + \widehat k)} \over 3}$
$\therefore$ $((\alpha + 4)\widehat i + 2\widehat j + 3\widehat k).{{(2\widehat i + 2\widehat j + \widehat k)} \over 3} = 9$
or $\alpha$ = 6
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
Let $\overrightarrow x $ be a vector in the plane containing vectors $\overrightarrow a = 2\widehat i - \widehat j + \widehat k$ and $\overrightarrow b = \widehat i + 2\widehat j - \widehat k$. If the vector $\overrightarrow x $ is perpendicular to $\left( {3\widehat i + 2\widehat j - \widehat k} \right)$ and its projection on $\overrightarrow a $ is ${{17\sqrt 6 } \over 2}$, then the value of $|\overrightarrow x {|^2}$ is equal to __________.
Correct Answer: 486
Explanation:
Let, $\overrightarrow x = k(\overrightarrow a + \lambda \overrightarrow b )$
$\overrightarrow x$ is perpendicular to $3\widehat i + 2\widehat j - \widehat k$
I. k{(2 + $\lambda$)3 + (2$\lambda$ $-$ 1)2 + (1 $-$ $\lambda$)($-$1) = 0
$ \Rightarrow $ 8$\lambda$ + 3 = 0
$\lambda = {{ - 3} \over 8}$
II. Also projection of $\overrightarrow x $ on $\overrightarrow a $ is ${{17\sqrt 6 } \over 2}$ therefore
${{\overrightarrow x .\overrightarrow a } \over {|\overrightarrow a |}} = {{17\sqrt 6 } \over 2}$
$ \Rightarrow k\left\{ {{{(\overrightarrow a + \lambda \overrightarrow b ).\overrightarrow a } \over {\sqrt 6 }}} \right\} = {{17\sqrt 6 } \over 2}$
$ \Rightarrow k\left\{ {6 + \left( {{3 \over 8}} \right)} \right\} = {{17 \times 6} \over 2}$
$ \Rightarrow k = {{51} \over {51}} \times 8$
k = 8
$ \therefore $ $\overrightarrow x = 8\left( {{{13} \over 8}\widehat i - {{14} \over 8}\widehat j + {{11} \over 8}\widehat k} \right)$
$ = 13\widehat i - 14\widehat j + 11\widehat k$
$|\overrightarrow x {|^2} = 169 + 196 + 121 = 486$
$\overrightarrow x$ is perpendicular to $3\widehat i + 2\widehat j - \widehat k$
I. k{(2 + $\lambda$)3 + (2$\lambda$ $-$ 1)2 + (1 $-$ $\lambda$)($-$1) = 0
$ \Rightarrow $ 8$\lambda$ + 3 = 0
$\lambda = {{ - 3} \over 8}$
II. Also projection of $\overrightarrow x $ on $\overrightarrow a $ is ${{17\sqrt 6 } \over 2}$ therefore
${{\overrightarrow x .\overrightarrow a } \over {|\overrightarrow a |}} = {{17\sqrt 6 } \over 2}$
$ \Rightarrow k\left\{ {{{(\overrightarrow a + \lambda \overrightarrow b ).\overrightarrow a } \over {\sqrt 6 }}} \right\} = {{17\sqrt 6 } \over 2}$
$ \Rightarrow k\left\{ {6 + \left( {{3 \over 8}} \right)} \right\} = {{17 \times 6} \over 2}$
$ \Rightarrow k = {{51} \over {51}} \times 8$
k = 8
$ \therefore $ $\overrightarrow x = 8\left( {{{13} \over 8}\widehat i - {{14} \over 8}\widehat j + {{11} \over 8}\widehat k} \right)$
$ = 13\widehat i - 14\widehat j + 11\widehat k$
$|\overrightarrow x {|^2} = 169 + 196 + 121 = 486$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
If $\overrightarrow a = \alpha \widehat i + \beta \widehat j + 3\widehat k$,
$\overrightarrow b = - \beta \widehat i - \alpha \widehat j - \widehat k$ and
$\overrightarrow c = \widehat i - 2\widehat j - \widehat k$
such that $\overrightarrow a \,.\,\overrightarrow b = 1$ and $\overrightarrow b \,.\,\overrightarrow c = - 3$, then ${1 \over 3}\left( {\left( {\overrightarrow a \times \overrightarrow b } \right)\,.\,\overrightarrow c } \right)$ is equal to _____________.
$\overrightarrow b = - \beta \widehat i - \alpha \widehat j - \widehat k$ and
$\overrightarrow c = \widehat i - 2\widehat j - \widehat k$
such that $\overrightarrow a \,.\,\overrightarrow b = 1$ and $\overrightarrow b \,.\,\overrightarrow c = - 3$, then ${1 \over 3}\left( {\left( {\overrightarrow a \times \overrightarrow b } \right)\,.\,\overrightarrow c } \right)$ is equal to _____________.
Correct Answer: 2
Explanation:
$\overrightarrow a .\overrightarrow b = 1 \Rightarrow - \alpha \beta - \alpha \beta - 3 = 1$
$ \Rightarrow \alpha \beta = - 2$ .... (i)
$\overrightarrow b .\overrightarrow c = - 3 \Rightarrow - \beta + 2\alpha + 1 = - 3$
$2\alpha - \beta = - 4$ ..... (ii)
Solving (i) & (ii) $\alpha$ = $-$1, $\beta$ = 2,
${1 \over 3}((\overrightarrow a \, \times \overrightarrow b )\,.\,\overrightarrow c ) = {1 \over 3}\left| {\matrix{ { - 1} & 2 & 3 \cr { - 2} & 1 & { - 1} \cr 1 & { - 2} & { - 1} \cr } } \right| = 2$
$ \Rightarrow \alpha \beta = - 2$ .... (i)
$\overrightarrow b .\overrightarrow c = - 3 \Rightarrow - \beta + 2\alpha + 1 = - 3$
$2\alpha - \beta = - 4$ ..... (ii)
Solving (i) & (ii) $\alpha$ = $-$1, $\beta$ = 2,
${1 \over 3}((\overrightarrow a \, \times \overrightarrow b )\,.\,\overrightarrow c ) = {1 \over 3}\left| {\matrix{ { - 1} & 2 & 3 \cr { - 2} & 1 & { - 1} \cr 1 & { - 2} & { - 1} \cr } } \right| = 2$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
Let $\overrightarrow c $ be a vector perpendicular to the vectors, $\overrightarrow a $ = $\widehat i$ + $\widehat j$ $-$ $\widehat k$ and
$\overrightarrow b $ = $\widehat i$ + 2$\widehat j$ + $\widehat k$. If $\overrightarrow c \,.\,\left( {\widehat i + \widehat j + 3\widehat k} \right)$ = 8 then the value of
$\overrightarrow c $ . $\left( {\overrightarrow a \times \overrightarrow b } \right)$ is equal to __________.
$\overrightarrow b $ = $\widehat i$ + 2$\widehat j$ + $\widehat k$. If $\overrightarrow c \,.\,\left( {\widehat i + \widehat j + 3\widehat k} \right)$ = 8 then the value of
$\overrightarrow c $ . $\left( {\overrightarrow a \times \overrightarrow b } \right)$ is equal to __________.
Correct Answer: 28
Explanation:
$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 1 & { - 1} \cr
1 & 2 & 1 \cr
} } \right| = (3, - 2,1)$
$\overrightarrow c \bot \overrightarrow a ,\overrightarrow c \bot \overrightarrow b \Rightarrow C||\overrightarrow a \times \overrightarrow b $
$\overrightarrow c = \lambda (\overrightarrow a \times \overrightarrow b )$
$ \Rightarrow \overrightarrow c = \lambda (3\widehat i - 2\widehat j + \widehat k)$
Given, $\overrightarrow c .(\widehat i + \widehat j + 3\widehat k) = 8$
$ \Rightarrow 3\lambda - 2\lambda + 3\lambda = 8$
$ \Rightarrow 4\lambda = 8 \Rightarrow \lambda = 2$
$ \therefore $ $\overrightarrow c = 6\widehat i - 4\widehat j + 2\widehat k$
$\overrightarrow c \,.\,(\overrightarrow a \times \overrightarrow b ) = [\overrightarrow c \overrightarrow a \overrightarrow b ] = \left| {\matrix{ 6 & { - 4} & 2 \cr 1 & 1 & { - 1} \cr 1 & 2 & 1 \cr } } \right|$
$ \Rightarrow $ 18 + 8 + 2 = 28
$\overrightarrow c \bot \overrightarrow a ,\overrightarrow c \bot \overrightarrow b \Rightarrow C||\overrightarrow a \times \overrightarrow b $
$\overrightarrow c = \lambda (\overrightarrow a \times \overrightarrow b )$
$ \Rightarrow \overrightarrow c = \lambda (3\widehat i - 2\widehat j + \widehat k)$
Given, $\overrightarrow c .(\widehat i + \widehat j + 3\widehat k) = 8$
$ \Rightarrow 3\lambda - 2\lambda + 3\lambda = 8$
$ \Rightarrow 4\lambda = 8 \Rightarrow \lambda = 2$
$ \therefore $ $\overrightarrow c = 6\widehat i - 4\widehat j + 2\widehat k$
$\overrightarrow c \,.\,(\overrightarrow a \times \overrightarrow b ) = [\overrightarrow c \overrightarrow a \overrightarrow b ] = \left| {\matrix{ 6 & { - 4} & 2 \cr 1 & 1 & { - 1} \cr 1 & 2 & 1 \cr } } \right|$
$ \Rightarrow $ 18 + 8 + 2 = 28
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
Let $\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$ and $\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\overrightarrow a $ and $\overrightarrow b $ is $8\sqrt 3 $ square units, then $\overrightarrow a $ . $\overrightarrow b $ is equal to __________.
Correct Answer: 2
Explanation:
$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$
$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$
Area of parallelogram = $\left| {\overrightarrow a \times \overrightarrow b } \right|$
$ = \left| {(\widehat i + \alpha \widehat j + 3\widehat k) \times (3\widehat i - \alpha \widehat j + \widehat k)} \right|$
$8\sqrt 3 = \left| {(4\alpha )\widehat i + 8\widehat j - (4\alpha )\widehat k} \right|$
$(64)(3) = 16{\alpha ^2} + 64 + 16{\alpha ^2}$
$(64)(3) = 32{\alpha ^2} + 64$
$6 = {\alpha ^2} + 2$
${\alpha ^2} = 4$
$ \therefore $ $\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$
$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$
$\overrightarrow a \,.\,\overrightarrow b = 3 - {\alpha ^2} + 3$
$ = 6 - {\alpha ^2}$
$ = 6 - 4$
$ = 2$
$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$
Area of parallelogram = $\left| {\overrightarrow a \times \overrightarrow b } \right|$
$ = \left| {(\widehat i + \alpha \widehat j + 3\widehat k) \times (3\widehat i - \alpha \widehat j + \widehat k)} \right|$
$8\sqrt 3 = \left| {(4\alpha )\widehat i + 8\widehat j - (4\alpha )\widehat k} \right|$
$(64)(3) = 16{\alpha ^2} + 64 + 16{\alpha ^2}$
$(64)(3) = 32{\alpha ^2} + 64$
$6 = {\alpha ^2} + 2$
${\alpha ^2} = 4$
$ \therefore $ $\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$
$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$
$\overrightarrow a \,.\,\overrightarrow b = 3 - {\alpha ^2} + 3$
$ = 6 - {\alpha ^2}$
$ = 6 - 4$
$ = 2$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
Let $\overrightarrow a = \widehat i + 2\widehat j - \widehat k$, $\overrightarrow b = \widehat i - \widehat j$ and $\overrightarrow c = \widehat i - \widehat j - \widehat k$ be three given vectors. If $\overrightarrow r $ is a vector such that $\overrightarrow r \times \overrightarrow a = \overrightarrow c \times \overrightarrow a $ and $\overrightarrow r .\,\overrightarrow b = 0$, then $\overrightarrow r .\,\overrightarrow a $ is equal to __________.
Correct Answer: 12
Explanation:
Given, $\overrightarrow a = \widehat i + 2\widehat j - \widehat k$,
$\overrightarrow b = \widehat i - \widehat j$,
$\overrightarrow c = \widehat i - \widehat j - \widehat k$
$\overrightarrow r \times \overrightarrow a = \overrightarrow c \times \overrightarrow a $
$ \Rightarrow \overrightarrow r \times \overrightarrow a - \overrightarrow c \times \overrightarrow a = 0$
$ \Rightarrow (\overrightarrow r - \overrightarrow c ) \times \overrightarrow a = 0$
$\therefore$ $\overrightarrow r - \overrightarrow c = \lambda \overrightarrow a $
$ \Rightarrow \overrightarrow r = \lambda \overrightarrow a + \overrightarrow c $
$ \Rightarrow \overrightarrow r \,.\,\overrightarrow b = \lambda \overrightarrow a \,.\,\overrightarrow b + \overrightarrow c \,.\,\overrightarrow b $ (taking dot with $\overrightarrow b $)
$ \Rightarrow 0 = \lambda \overrightarrow a \,.\,\overrightarrow b + \overrightarrow c \,.\,\overrightarrow b $ [$\because$ $\overrightarrow r \,.\,\overrightarrow b = 0$]
$ \Rightarrow \lambda (\widehat i + 2\widehat j - \widehat k)\,.\,(\widehat i - \widehat j) + (\widehat i - \widehat j - \widehat k)\,.\,(\widehat i - \widehat j) = 0$
$ \Rightarrow \lambda (1 - 2) + 2 = 0$
$ \Rightarrow \lambda = 2$
$\therefore$ $\overrightarrow r = 2\overrightarrow a + \overrightarrow c $
$ \Rightarrow \overrightarrow r \,.\,\overrightarrow a = 2\overrightarrow a \,.\,\overrightarrow a + \overrightarrow c \,.\,\overrightarrow a $ [taking dot with ${\overrightarrow a }$]
$ = 2{\left| {\overrightarrow a } \right|^2} + \overrightarrow a \,.\,\overrightarrow c $
$ = 2(1 + 4 + 1) + (1 - 2 + 1)$
$ \Rightarrow \overrightarrow r \,.\,\overrightarrow a = 12$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
Let three vectors $\overrightarrow a ,\overrightarrow b $ and $\overrightarrow c $ be such that $\overrightarrow c $ is coplanar
with $\overrightarrow a $ and $\overrightarrow b $, $\overrightarrow a .\overrightarrow c $ = 7 and $\overrightarrow b $ is perpendicular to $\overrightarrow c $, where
$\overrightarrow a = - \widehat i + \widehat j + \widehat k$ and $\overrightarrow b = 2\widehat i + \widehat k$ , then the
value of $2{\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2}$ is _____.
with $\overrightarrow a $ and $\overrightarrow b $, $\overrightarrow a .\overrightarrow c $ = 7 and $\overrightarrow b $ is perpendicular to $\overrightarrow c $, where
$\overrightarrow a = - \widehat i + \widehat j + \widehat k$ and $\overrightarrow b = 2\widehat i + \widehat k$ , then the
value of $2{\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2}$ is _____.
Correct Answer: 75
Explanation:
$\overrightarrow c = \lambda (\overrightarrow b \times (\overrightarrow a \times \overrightarrow b ))$
$ = \lambda ((\overrightarrow b \,.\,\overrightarrow b )\overrightarrow a - (\overrightarrow b \,.\,\overrightarrow a )\overrightarrow b )$
$ = \lambda (5( - \widehat i + \widehat j + \widehat k) + 2\widehat i + \widehat k)$
$ = \lambda ( - 3\widehat i + 5\widehat j + 6\widehat k)$
$\overrightarrow c \,.\,\overrightarrow a = 7 $
$\Rightarrow 3\lambda + 5\lambda + 6\lambda = 7$
$ \Rightarrow $ $\lambda = {1 \over 2}$
$ \therefore $ $2{\left| {\left( {{{ - 3} \over 2} - 1 + 2} \right)\widehat i + \left( {{5 \over 2} + 1} \right)\widehat j + (3 + 1 + 1)\widehat k} \right|^2}$
$ = 2\left( {{1 \over 4} + {{49} \over 4} + 25} \right) = 25 + 50 = 75$
$ = \lambda ((\overrightarrow b \,.\,\overrightarrow b )\overrightarrow a - (\overrightarrow b \,.\,\overrightarrow a )\overrightarrow b )$
$ = \lambda (5( - \widehat i + \widehat j + \widehat k) + 2\widehat i + \widehat k)$
$ = \lambda ( - 3\widehat i + 5\widehat j + 6\widehat k)$
$\overrightarrow c \,.\,\overrightarrow a = 7 $
$\Rightarrow 3\lambda + 5\lambda + 6\lambda = 7$
$ \Rightarrow $ $\lambda = {1 \over 2}$
$ \therefore $ $2{\left| {\left( {{{ - 3} \over 2} - 1 + 2} \right)\widehat i + \left( {{5 \over 2} + 1} \right)\widehat j + (3 + 1 + 1)\widehat k} \right|^2}$
$ = 2\left( {{1 \over 4} + {{49} \over 4} + 25} \right) = 25 + 50 = 75$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
If the volume of a parallelopiped, whose
coterminus edges are given by the
vectors $\overrightarrow a = \widehat i + \widehat j + n\widehat k$,
$\overrightarrow b = 2\widehat i + 4\widehat j - n\widehat k$ and
$\overrightarrow c = \widehat i + n\widehat j + 3\widehat k$ ($n \ge 0$), is 158 cu. units, then :
coterminus edges are given by the
vectors $\overrightarrow a = \widehat i + \widehat j + n\widehat k$,
$\overrightarrow b = 2\widehat i + 4\widehat j - n\widehat k$ and
$\overrightarrow c = \widehat i + n\widehat j + 3\widehat k$ ($n \ge 0$), is 158 cu. units, then :
A.
n = 7
B.
$\overrightarrow b .\overrightarrow c = 10$
C.
$\overrightarrow a .\overrightarrow c = 17$
D.
n = 9
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
Let x0 be the point of Local maxima of $f(x) = \overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)$, where
$\overrightarrow a = x\widehat i - 2\widehat j + 3\widehat k$, $\overrightarrow b = - 2\widehat i + x\widehat j - \widehat k$, $\overrightarrow c = 7\widehat i - 2\widehat j + x\widehat k$. Then the value of
$\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a $ at x = x0 is :
$\overrightarrow a = x\widehat i - 2\widehat j + 3\widehat k$, $\overrightarrow b = - 2\widehat i + x\widehat j - \widehat k$, $\overrightarrow c = 7\widehat i - 2\widehat j + x\widehat k$. Then the value of
$\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a $ at x = x0 is :
A.
14
B.
-30
C.
-4
D.
-22
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
Let a, b c $ \in $ R be such that a2
+ b2
+ c2
= 1. If
$a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right)$,
where ${\theta = {\pi \over 9}}$, then the angle between the vectors $a\widehat i + b\widehat j + c\widehat k$ and $b\widehat i + c\widehat j + a\widehat k$ is :
$a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right)$,
where ${\theta = {\pi \over 9}}$, then the angle between the vectors $a\widehat i + b\widehat j + c\widehat k$ and $b\widehat i + c\widehat j + a\widehat k$ is :
A.
0
B.
${{\pi \over 9}}$
C.
${{{2\pi } \over 3}}$
D.
${{\pi \over 2}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
The lines
$\overrightarrow r = \left( {\widehat i - \widehat j} \right) + l\left( {2\widehat i + \widehat k} \right)$ and
$\overrightarrow r = \left( {2\widehat i - \widehat j} \right) + m\left( {\widehat i + \widehat j + \widehat k} \right)$
$\overrightarrow r = \left( {\widehat i - \widehat j} \right) + l\left( {2\widehat i + \widehat k} \right)$ and
$\overrightarrow r = \left( {2\widehat i - \widehat j} \right) + m\left( {\widehat i + \widehat j + \widehat k} \right)$
A.
do not intersect for any values of $l$ and m
B.
intersect for all values of $l$ and m
C.
intersect when $l$ = 2 and m = ${1 \over 2}$
D.
intersect when $l$ = 1 and m = 2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
Let $\overrightarrow a = \widehat i - 2\widehat j + \widehat k$ and $\overrightarrow b = \widehat i - \widehat j + \widehat k$ be two
vectors. If $\overrightarrow c $ is a vector such that $\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a $ and $\overrightarrow c .\overrightarrow a = 0$, then $\overrightarrow c .\overrightarrow b $ is equal to
A.
$ - {1 \over 2}$
B.
$ - {3 \over 2}$
C.
${1 \over 2}$
D.
-1
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
Let the volume of a parallelopiped whose
coterminous edges are given by
$\overrightarrow u = \widehat i + \widehat j + \lambda \widehat k$, $\overrightarrow v = \widehat i + \widehat j + 3\widehat k$ and
$\overrightarrow w = 2\widehat i + \widehat j + \widehat k$ be 1 cu. unit. If $\theta $ be the angle between the edges $\overrightarrow u $ and $\overrightarrow w $ , then cos$\theta $ can be :
$\overrightarrow u = \widehat i + \widehat j + \lambda \widehat k$, $\overrightarrow v = \widehat i + \widehat j + 3\widehat k$ and
$\overrightarrow w = 2\widehat i + \widehat j + \widehat k$ be 1 cu. unit. If $\theta $ be the angle between the edges $\overrightarrow u $ and $\overrightarrow w $ , then cos$\theta $ can be :
A.
${7 \over {6\sqrt 3 }}$
B.
${7 \over {6\sqrt 6 }}$
C.
${5 \over 7}$
D.
${5 \over {3\sqrt 3 }}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
Let $\overrightarrow a $
, $\overrightarrow b $
and $\overrightarrow c $
be three unit vectors such that
$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $. If $\lambda = \overrightarrow a .\vec b + \vec b.\overrightarrow c + \overrightarrow c .\overrightarrow a $ and
$\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a $, then the ordered pair, $\left( {\lambda ,\overrightarrow d } \right)$ is equal to :
$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $. If $\lambda = \overrightarrow a .\vec b + \vec b.\overrightarrow c + \overrightarrow c .\overrightarrow a $ and
$\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a $, then the ordered pair, $\left( {\lambda ,\overrightarrow d } \right)$ is equal to :
A.
$\left( {{3 \over 2},3\overrightarrow a \times \overrightarrow c } \right)$
B.
$\left( { - {3 \over 2},3\overrightarrow c \times \overrightarrow b } \right)$
C.
$\left( { - {3 \over 2},3\overrightarrow a \times \overrightarrow b } \right)$
D.
$\left( {{3 \over 2},3\overrightarrow b \times \overrightarrow c } \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
A vector $\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k\left( {\alpha ,\beta \in R} \right)$ lies in the plane of the vectors, $\overrightarrow b = \widehat i + \widehat j$ and $\overrightarrow c = \widehat i - \widehat j + 4\widehat k$. If $\overrightarrow a $ bisects the angle between $\overrightarrow b $ and $\overrightarrow c $, then:
A.
$\overrightarrow a .\widehat i + 3 = 0$
B.
$\overrightarrow a .\widehat k - 4 = 0$
C.
$\overrightarrow a .\widehat i + 1 = 0$
D.
$\overrightarrow a .\widehat k + 2 = 0$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Evening Slot
If $\overrightarrow x $ and $\overrightarrow y $ be two non-zero vectors such that
$\left| {\overrightarrow x + \overrightarrow y } \right| = \left| {\overrightarrow x } \right|$ and ${2\overrightarrow x + \lambda \overrightarrow y }$ is perpendicular to ${\overrightarrow y }$,
then the value of $\lambda $ is _________ .
Correct Answer: 1
Explanation:
$\left| {\overrightarrow x + \overrightarrow y } \right| = \left| {\overrightarrow x } \right|$
Squaring both sides we get
${\left| {\overrightarrow x } \right|^2} + 2\overrightarrow x .\overrightarrow y + {\left| {\overrightarrow y } \right|^2} = {\left| {\overrightarrow x } \right|^2}$
$ \Rightarrow $ $2\overrightarrow x .\overrightarrow y + \overrightarrow y .\overrightarrow y $ = 0 ....(1)
Given ${2\overrightarrow x + \lambda \overrightarrow y }$ is perpendicular to ${\overrightarrow y }$
$ \therefore $ $\left( {2\overrightarrow x + \lambda \overrightarrow y } \right).\overrightarrow y $ = 0
$ \Rightarrow $ $2\overrightarrow x .\overrightarrow y + \lambda \overrightarrow y .\overrightarrow y $ = 0 ....(2)
Comparing (1) & (2) we get, $\lambda $ = 1
Squaring both sides we get
${\left| {\overrightarrow x } \right|^2} + 2\overrightarrow x .\overrightarrow y + {\left| {\overrightarrow y } \right|^2} = {\left| {\overrightarrow x } \right|^2}$
$ \Rightarrow $ $2\overrightarrow x .\overrightarrow y + \overrightarrow y .\overrightarrow y $ = 0 ....(1)
Given ${2\overrightarrow x + \lambda \overrightarrow y }$ is perpendicular to ${\overrightarrow y }$
$ \therefore $ $\left( {2\overrightarrow x + \lambda \overrightarrow y } \right).\overrightarrow y $ = 0
$ \Rightarrow $ $2\overrightarrow x .\overrightarrow y + \lambda \overrightarrow y .\overrightarrow y $ = 0 ....(2)
Comparing (1) & (2) we get, $\lambda $ = 1
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Morning Slot
If $\overrightarrow a $
and $\overrightarrow b $
are unit vectors, then the greatest value of
$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$ is_____.
$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$ is_____.
Correct Answer: 4
Explanation:
Let angle between $\overrightarrow a $ and $\overrightarrow b $
be $\theta $.
$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$
= $\sqrt 3 \left( {\sqrt {1 + 1 + 2\cos \theta } } \right)$ + $\left( {\sqrt {1 + 1 - 2\cos \theta } } \right)$
= $\sqrt 3 \left( {\sqrt {2 + 2\cos \theta } } \right)$ + $\left( {\sqrt {2 - 2\cos \theta } } \right)$
= $\sqrt 6 \left( {\sqrt {1 + \cos \theta } } \right)$ + $\sqrt 2 \left( {\sqrt {1 - \cos \theta } } \right)$
= $\sqrt 6 \left( {\sqrt {2{{\cos }^2}{\theta \over 2}} } \right)$ + $\sqrt 2 \left( {\sqrt {2{{\sin }^2}{\theta \over 2}} } \right)$
= $2\sqrt 3 \left| {\cos {\theta \over 2}} \right|$ + 2$\left| {\sin {\theta \over 2}} \right|$
$ \le $ $\sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2}} $ = 4
Note : |x| = $\sqrt {{x^2}} $
|x - 1| = $\sqrt {{{\left( {x - 1} \right)}^2}} $
|sin x| = $\sqrt {{{\sin }^2}x} $
That is why ${\sqrt {{{\sin }^2}{\theta \over 2}} }$ = $\left| {\sin {\theta \over 2}} \right|$ and ${\sqrt {{{\cos }^2}{\theta \over 2}} }$ = $\left| {\cos {\theta \over 2}} \right|$
$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$
= $\sqrt 3 \left( {\sqrt {1 + 1 + 2\cos \theta } } \right)$ + $\left( {\sqrt {1 + 1 - 2\cos \theta } } \right)$
= $\sqrt 3 \left( {\sqrt {2 + 2\cos \theta } } \right)$ + $\left( {\sqrt {2 - 2\cos \theta } } \right)$
= $\sqrt 6 \left( {\sqrt {1 + \cos \theta } } \right)$ + $\sqrt 2 \left( {\sqrt {1 - \cos \theta } } \right)$
= $\sqrt 6 \left( {\sqrt {2{{\cos }^2}{\theta \over 2}} } \right)$ + $\sqrt 2 \left( {\sqrt {2{{\sin }^2}{\theta \over 2}} } \right)$
= $2\sqrt 3 \left| {\cos {\theta \over 2}} \right|$ + 2$\left| {\sin {\theta \over 2}} \right|$
$ \le $ $\sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2}} $ = 4
Note : |x| = $\sqrt {{x^2}} $
|x - 1| = $\sqrt {{{\left( {x - 1} \right)}^2}} $
|sin x| = $\sqrt {{{\sin }^2}x} $
That is why ${\sqrt {{{\sin }^2}{\theta \over 2}} }$ = $\left| {\sin {\theta \over 2}} \right|$ and ${\sqrt {{{\cos }^2}{\theta \over 2}} }$ = $\left| {\cos {\theta \over 2}} \right|$