iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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 :
A.
14
B.
-30
C.
-4
D.
-22
Correct Answer: D
Explanation:
$f(x) = \overrightarrow a \,.\,(\overrightarrow b \times \overrightarrow c )$
$ = \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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.
0
B.
${{\pi \over 9}}$
C.
${{{2\pi } \over 3}}$
D.
${{\pi \over 2}}$
Correct Answer: D
Explanation:
Let, $\overrightarrow {{a_1}} = a\widehat i + b\widehat j + c\widehat k$
and $\overrightarrow {{a_2}} = b\widehat i + c\widehat j + a\widehat k$
Equating coefficient of $\widehat i$, $\widehat j$ and $\widehat k$ of L1
and L2
2l + 1 = m + 2 ... (1)
–1 = –1 + m ...(2)
l = –m ...(3)
from (ii) m = 0
from (iii) $l$ = 0
These values of m and $l$ do not satisfy equation (1).
Hence the two lines do not intersect for any values of $l$ and m.
2020
Q204
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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
Correct Answer: A
Explanation:
$\overrightarrow a = \widehat i - 2\widehat j + \widehat k$
$\overrightarrow b = \widehat i - \widehat j + \widehat k$
$\left| {\overrightarrow a } \right|$ = $\sqrt 6 $, $\left| {\overrightarrow b } \right|$ = $\sqrt 3 $
and ${\overrightarrow a .\overrightarrow b }$ = 4
Given $\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a $
$ \Rightarrow $ ${\left( {\overrightarrow b \times \overrightarrow c } \right)}$ - ${\left( {\overrightarrow b \times \overrightarrow a } \right)}$ = 0
$ \Rightarrow $ ${\overrightarrow b \times \left( {\overrightarrow c - \overrightarrow a } \right)}$ = 0
$ \therefore $ ${\overrightarrow b \parallel \left( {\overrightarrow c - \overrightarrow a } \right)}$
$ \Rightarrow $ ${\left( {\overrightarrow c - \overrightarrow a } \right) = \lambda \overrightarrow b }$
$ \Rightarrow $ ${\overrightarrow c = \overrightarrow a + \lambda \overrightarrow b }$
$ \Rightarrow $ ${\overrightarrow c .\overrightarrow a = \overrightarrow a .\overrightarrow a + \lambda \overrightarrow a .\overrightarrow b }$
$ \Rightarrow $ 0 = ${{{\left| {\overrightarrow a } \right|}^2} + \lambda \left( {\overrightarrow a .\overrightarrow b } \right)}$
$ \Rightarrow $ $\lambda $ = ${{{ - {{\left| {\overrightarrow a } \right|}^2}} \over {\overrightarrow a .\overrightarrow b }}}$ = ${{ - 6} \over 4} = - {3 \over 2}$
$ \therefore $ $\overrightarrow c $ = ${\overrightarrow a - {3 \over 2}\overrightarrow b }$
$ \Rightarrow $ $\overrightarrow c $ = ($\widehat i - 2\widehat j + \widehat k$) - ${3 \over 2}$($\widehat i - \widehat j + \widehat k$)
$ \therefore $ $\overrightarrow c .\overrightarrow b $ = $ - {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)$($\widehat i - \widehat j + \widehat k$)
$ \therefore $ $\overrightarrow c .\overrightarrow b $ = $ - {1 \over 2}$
2020
Q205
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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 :
A.
${7 \over {6\sqrt 3 }}$
B.
${7 \over {6\sqrt 6 }}$
C.
${5 \over 7}$
D.
${5 \over {3\sqrt 3 }}$
Correct Answer: A
Explanation:
Volume of parallelopiped = 1
$\left| {\left[ {\matrix{
{\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr
} } \right]} \right|$ = 1
$\overrightarrow u = \widehat i + \widehat j + 2 \widehat k$ or
$\overrightarrow u = \widehat i + \widehat j + 4 \widehat k$
$ \therefore $ cos $\theta $ = ${{{\overrightarrow u .\overrightarrow w } \over {\left| {\overrightarrow u } \right|\left| {\overrightarrow w } \right|}}}$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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 :
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.
$\left( {{3 \over 2},3\overrightarrow b \times \overrightarrow c } \right)$
Correct Answer: C
Explanation:
$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $
$ \Rightarrow $ ${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2}$ = 0
$ \Rightarrow $ ${{{\left| {\overrightarrow a } \right|}^2}}$ + ${{{\left| {\overrightarrow b } \right|}^2}}$ + ${{{\left| {\overrightarrow c } \right|}^2}}$ + ${2\left( {\overrightarrow a .\overrightarrow b } \right)}$ + ${2\left( {\overrightarrow b .\overrightarrow c } \right)}$ + ${2\left( {\overrightarrow c .\overrightarrow {a} } \right)}$ = 0
$ \Rightarrow $ 3 + ${2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right)}$ = 0
$ \Rightarrow $ 3 + 2$\lambda $ = 0
$ \Rightarrow $ $\lambda $ = $ - {3 \over 2}$
$\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a $
= ${\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \left( { - \overrightarrow a - \overrightarrow b } \right) + \left( { - \overrightarrow a - \overrightarrow b } \right) \times \overrightarrow a }$
= ${\overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow a + }$ ${\overrightarrow b \times \overrightarrow b - \overrightarrow a \times \overrightarrow a - \overrightarrow b \times \overrightarrow a }$
= ${\overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b }$
= ${3\left( {\overrightarrow a \times \overrightarrow b } \right)}$
2020
Q207
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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.
$\overrightarrow a .\widehat k + 2 = 0$
Correct Answer: B
Explanation:
Angle bisector $\overrightarrow a = \lambda \left( {\widehat b + \widehat c} \right)$
$ \Rightarrow $ $\overrightarrow a = {\lambda \over {3\sqrt 2 }}\left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$
comparing with $\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k$
${{2\lambda } \over {3\sqrt 2 }}$ = 2
$ \Rightarrow $ $\lambda $ = ${3\sqrt 2 }$
$ \therefore $ $\overrightarrow a = \left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$
Then $\overrightarrow a .\widehat k - 4 $
= 4 - 4 = 0
2020
Q208
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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
2020
Q209
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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_____.
Correct Answer: 4
Explanation:
Let angle between $\overrightarrow a $ and $\overrightarrow b $
be $\theta $.
$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$
That is why ${\sqrt {{{\sin }^2}{\theta \over 2}} }$ = $\left| {\sin {\theta \over 2}} \right|$ and ${\sqrt {{{\cos }^2}{\theta \over 2}} }$ = $\left| {\cos {\theta \over 2}} \right|$
2020
Q210
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let the vectors $\overrightarrow a $, $\overrightarrow b $, $\overrightarrow c $
be such that
$\left| {\overrightarrow a } \right| = 2$, $\left| {\overrightarrow b } \right| = 4$
and $\left| {\overrightarrow c } \right| = 4$. If the projection of
$\overrightarrow b $
on $\overrightarrow a $
is equal to the projection of $\overrightarrow c $
on $\overrightarrow a $
and $\overrightarrow b $
is perpendicular to $\overrightarrow c $,
then the value of
$\left| {\overrightarrow a + \vec b - \overrightarrow c } \right|$
is ___________.
Correct Answer: 6
Explanation:
Projection of $\overrightarrow b $
on $\overrightarrow a $
= Projection of $\overrightarrow c $
on $\overrightarrow a $
$ \Rightarrow $ ${{\overrightarrow b .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = {{\overrightarrow c .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}}$
$ \Rightarrow $ $\overrightarrow b .\overrightarrow a = \overrightarrow c .\overrightarrow a $
$ \because $ $\overrightarrow b $
is perpendicular to $\overrightarrow c $
$ \therefore $ $\overrightarrow b .\overrightarrow c = 0$
Let $\left| {\overrightarrow a + \vec b - \overrightarrow c } \right|$ = k
Square both sides
k2 = ${{{\left( {\overrightarrow a } \right)}^2}}$ + ${{{\left( {\overrightarrow b } \right)}^2}}$ + ${{{\left( {\overrightarrow c } \right)}^2}}$ + $2\overrightarrow a .\overrightarrow b $ - $2\overrightarrow b .\overrightarrow c $ - $2\overrightarrow a .\overrightarrow c $
$ \Rightarrow $ k2 = ${{{\left( {\overrightarrow a } \right)}^2}}$ + ${{{\left( {\overrightarrow b } \right)}^2}}$ + ${{{\left( {\overrightarrow c } \right)}^2}}$
$ \Rightarrow $ k2 = 22 + 42 + 42 = 36
$ \Rightarrow $ k = 6
2020
Q211
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If $\overrightarrow a = 2\widehat i + \widehat j + 2\widehat k$, then the value of
${\left| {\widehat i \times \left( {\overrightarrow a \times \widehat i} \right)} \right|^2} + {\left| {\widehat j \times \left( {\overrightarrow a \times \widehat j} \right)} \right|^2} + {\left| {\widehat k \times \left( {\overrightarrow a \times \widehat k} \right)} \right|^2}$ is equal to____
Correct Answer: 18
Explanation:
Let $\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$
Now $\widehat i \times \left( {\overrightarrow a \times \widehat i} \right) = \left( {\widehat i.\widehat i} \right)\overrightarrow a - \left( {\widehat i.\overrightarrow a } \right)\widehat i$
= $y\widehat j + z\widehat k$
Similarly $\widehat j \times \left( {\overrightarrow a \times \widehat j} \right) = x\widehat i + z\widehat k$
$\widehat k \times \left( {\overrightarrow a \times \widehat k} \right) = x\widehat i + y\widehat j$
Now ${\left| {y\widehat j + z\widehat k} \right|^2} + {\left| {x\widehat i + z\widehat k} \right|^2} + {\left| {x\widehat i + y\widehat j} \right|^2}$
= $2({x^2} + {y^2} + {z^2}) $
Given $\overrightarrow a = 2\widehat i + \widehat j + 2\widehat k$
$ \therefore $ x = 2, y = 1, z = 2
= 2(4 + 1 + 4) = 18
2020
Q212
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let the position vectors of points 'A' and 'B' be
$\widehat i + \widehat j + \widehat k$ and $2\widehat i + \widehat j + 3\widehat k$, respectively. A point
'P' divides the line segment AB internally in the
ratio
$\lambda $ : 1 (
$\lambda $ > 0). If O is the origin and
$\overrightarrow {OB} .\overrightarrow {OP} - 3{\left| {\overrightarrow {OA} \times \overrightarrow {OP} } \right|^2} = 6$, then
$\lambda $ is equal
to______.
Correct Answer: 0.8
Explanation:
Let, $\overrightarrow a $ = $\widehat i + \widehat j + \widehat k$
and $\overrightarrow b $ = $2\widehat i + \widehat j + 3\widehat k$
$\overrightarrow {OB} = \overrightarrow b $
$\overrightarrow {OP} = {{\overrightarrow a + \lambda \overrightarrow b } \over {1 + \lambda }}$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ be three unit vectors such that
${\left| {\overrightarrow a - \overrightarrow b } \right|^2}$ + ${\left| {\overrightarrow a - \overrightarrow c } \right|^2}$ = 8.
Then ${\left| {\overrightarrow a + 2\overrightarrow b } \right|^2}$ + ${\left| {\overrightarrow a + 2\overrightarrow c } \right|^2}$ is equal to ______.
Correct Answer: 2
Explanation:
Given, $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = 1$
${\left| {\overrightarrow a - \overrightarrow b } \right|^2}$ + ${\left| {\overrightarrow a - \overrightarrow c } \right|^2}$ = 8
$ \Rightarrow $ ${\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} - 2\overrightarrow a .\overrightarrow b + {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow c } \right|^2} - 2\overrightarrow a .\overrightarrow c $ = 8
$ \Rightarrow $ $\overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c $ = -2
Now, ${\left| {\overrightarrow a + 2\overrightarrow b } \right|^2}$ + ${\left| {\overrightarrow a + 2\overrightarrow c } \right|^2}$
= ${\left| {\overrightarrow a } \right|^2} + 4{\left| {\overrightarrow b } \right|^2} + 4\overrightarrow a .\overrightarrow b + {\left| {\overrightarrow a } \right|^2} + 4{\left| {\overrightarrow c } \right|^2} + 4\overrightarrow a .\overrightarrow c $
= 10 + 4$\left( {\overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c } \right)$
= 10 + 4(-2)
= 2
2020
Q214
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ be three vectors such that $\left| {\overrightarrow a } \right| = \sqrt 3 $,
$\left| {\overrightarrow b } \right| = 5,\overrightarrow b .\overrightarrow c = 10$ and the angle between $\overrightarrow b $ and $\overrightarrow c $
is ${\pi \over 3}$. If ${\overrightarrow a }$ is perpendicular to the vector $\overrightarrow b \times \overrightarrow c $ , then $\left| {\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)} \right|$ is equal to _____.
Correct Answer: 30
Explanation:
Given $\left| {\overrightarrow a } \right| = \sqrt 3 $,
$\left| {\overrightarrow b } \right| = 5$
Given $\overrightarrow b .\overrightarrow c = 10$
And the angle between $\overrightarrow b $ and $\overrightarrow c $
is ${\pi \over 3}$
$ \therefore $ $bc\cos {\pi \over 3}$ = 10
$ \Rightarrow $ c = 4
${\overrightarrow a }$ is perpendicular to the vector $\overrightarrow b \times \overrightarrow c $
$ \therefore $ $\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)$ = 0 and angle between them is ${\pi \over 2}$
Now $\left| {\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)} \right|$
= $\left| {\overrightarrow a } \right|\left| {\overrightarrow b \times \overrightarrow c } \right|\sin {\pi \over 2}$
= $\left| {\overrightarrow a } \right|$.${\left| {\overrightarrow b } \right|.\left| {\overrightarrow c } \right|}$$\sin {\pi \over 3}$.1
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If the vectors, $\overrightarrow p = \left( {a + 1} \right)\widehat i + a\widehat j + a\widehat k$,
$\overrightarrow q = a\widehat i + \left( {a + 1} \right)\widehat j + a\widehat k$ and
$\overrightarrow r = a\widehat i + a\widehat j + \left( {a + 1} \right)\widehat k\left( {a \in R} \right)$
are coplanar
and $3{\left( {\overrightarrow p .\overrightarrow q } \right)^2} - \lambda \left| {\overrightarrow r \times \overrightarrow q } \right|^2 = 0$, then the value of $\lambda $ is ______.
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a = 3\widehat i + 2\widehat j + 2\widehat k$ and $\overrightarrow b = \widehat i + 2\widehat j - 2\widehat k$ be two vectors. If a vector perpendicular to both the vectors
$\overrightarrow a + \overrightarrow b $ and $\overrightarrow a - \overrightarrow b $ has the magnitude 12 then one such vector is :
A.
$4\left( {2\widehat i - 2\widehat j - \widehat k} \right)$
$ \Rightarrow \overrightarrow r = \pm 4\left( {2\widehat i - 2\widehat j - \widehat k} \right)$
2019
Q218
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If the volume of parallelopiped formed by the vectors $\widehat i + \lambda \widehat j + \widehat k$, $\widehat j + \lambda \widehat k$ and $\lambda \widehat i + \widehat k$ is minimum, then $\lambda $ is
equal to :
Whose minimum value occur at $\lambda $ = ${1 \over {\sqrt 3 }}$
2019
Q219
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The distance of the point having position vector $ - \widehat i + 2\widehat j + 6\widehat k$
from the straight line passing through the point
(2, 3, – 4) and parallel to the vector, $6\widehat i + 3\widehat j - 4\widehat k$ is :
A.
6
B.
7
C.
$2\sqrt {13} $
D.
$4\sqrt 3 $
Correct Answer: B
Explanation:
$AD = \left| {{{\overrightarrow {AP} .\overrightarrow n } \over {\left| {\overrightarrow n } \right|}}} \right| = \sqrt {61} $
$ \Rightarrow PD = \sqrt {A{P^2} - A{D^2}} $
$ = \sqrt {110 - 61} $
= 7
2019
Q220
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let A (3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G
divides BM in the ratio, 2 : 1, then cos ($\angle $GOA) (O being the origin) is equal to :
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If a unit vector $\overrightarrow a $ makes angles $\pi $/3 with $\widehat i$ , $\pi $/ 4
with $\widehat j$ and $\theta $$ \in $(0, $\pi $) with $\widehat k$, then a value of $\theta $
is :-
A.
${{5\pi } \over {6}}$
B.
${{5\pi } \over {12}}$
C.
${{2\pi } \over {3}}$
D.
${{\pi } \over {4}}$
Correct Answer: C
Explanation:
A unit vector $\overrightarrow a $ makes angles $\pi $/3 with $\widehat i$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow \alpha = 3\widehat i + \widehat j$ and $\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$
. If $\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}} $,
where ${\overrightarrow \beta _1}$
is parallel to $\overrightarrow \alpha $ and $\overrightarrow {{\beta _2}} $
is perpendicular
to $\overrightarrow \alpha $ , then ${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} $
is equal to
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ be three unit vectors, out of which vectors $\overrightarrow b $ and $\overrightarrow c $ are non-parallel. If $\alpha $ and $\beta $ are the angles which vector $\overrightarrow a $ makes with vectors $\overrightarrow b $ and $\overrightarrow c $ respectively and $\overrightarrow a $ $ \times $ ($\overrightarrow b $ $ \times $ $\overrightarrow c $) = ${1 \over 2}\overrightarrow b $, then $\left| {\alpha - \beta } \right|$ is equal to :
A.
90o
B.
30o
C.
45o
D.
60o
Correct Answer: B
Explanation:
$\left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right).\overrightarrow c = {1 \over 2}\overrightarrow b $
$ \because $ $\overrightarrow b \,\,$ & $\overrightarrow c \,\,$ are linearly independent
$ \therefore $ $\overrightarrow a \,$.$\overrightarrow c \,$ = ${1 \over 2}$ & $\overrightarrow a .\overrightarrow b $ = 0
(All given vectors are unit vectors)
$ \therefore $ $\overrightarrow a $^$\overrightarrow c $ = 60o & $\overrightarrow a $^$\overrightarrow b $ = 90o
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The sum of the distinct real values of $\mu $, for which the vectors, $\mu \widehat i + \widehat j + \widehat k,$ $\widehat i + \mu \widehat j + \widehat k,$ $\widehat i + \widehat j + \mu \widehat k$ are co-planar, is :
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\sqrt 3 \widehat i + \widehat j,$ $\widehat i + \sqrt 3 \widehat j$ and $\beta \widehat i + \left( {1 - \beta } \right)\widehat j$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is ${3 \over {\sqrt 2 }}$, then the sum of all possible values of $\beta $ is :
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a = \widehat i + 2\widehat j + 4\widehat k,$ $\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k$ and $\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k$ be coplanar vectors. Then the non-zero vector $\overrightarrow a \times \overrightarrow c $ is :
A.
$ - 10\widehat i - 5\widehat j$
B.
$ - 10\widehat i + 5\widehat j$
C.
$ - 14\widehat i + 5\widehat j$
D.
$ - 14\widehat i - 5\widehat j$
Correct Answer: B
Explanation:
$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 0$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If $\overrightarrow \alpha $ = $\left( {\lambda - 2} \right)\overrightarrow a + \overrightarrow b $ and $\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow a + 3\overrightarrow b $ be two given vectors $\overrightarrow a $ and $\overrightarrow b $ are non-collinear. The value of $\lambda $ for which vectors $\overrightarrow \alpha $ and $\overrightarrow \beta $ are collinear, is -
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a = 2\widehat i + {\lambda _1}\widehat j + 3\widehat k,\,\,$ $\overrightarrow b = 4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k,$ and $\overrightarrow c = 3\widehat i + 6\widehat j + \left( {{\lambda _3} - 1} \right)\widehat k$ be three vectors such that $\overrightarrow b = 2\overrightarrow a $ and $\overrightarrow a $ is perpendicular to $\overrightarrow c $. Then a possible value of $\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right)$ is :
A.
(1, 5, 1)
B.
(1, 3, 1)
C.
$\left( { - {1 \over 2},4,0} \right)$
D.
$\left( {{1 \over 2},4, - 2} \right)$
Correct Answer: C
Explanation:
Given $\overrightarrow b = 2\overrightarrow a $
$ \therefore $ $4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k = 4\widehat i + 2{\lambda _1}\widehat j + 6\widehat k$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a = \widehat i + \widehat j + \sqrt 2 \widehat k,$ $\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + \sqrt 2 \widehat k$, $\overrightarrow c = 5\widehat i + \widehat j + \sqrt 2 \widehat k$ be three vectors such that the projection vector of $\overrightarrow b $ on $\overrightarrow a $ is $\overrightarrow a $.
If $\overrightarrow a + \overrightarrow b $ is perpendicular to $\overrightarrow c $ , then $\left| {\overrightarrow b } \right|$ is equal to :
A.
$\sqrt {32} $
B.
6
C.
$\sqrt {22} $
D.
4
Correct Answer: B
Explanation:
Projection of $\overrightarrow b $ on $\overrightarrow a $ is $\overrightarrow a $
$ \therefore $ ${{\overrightarrow b \cdot \overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right|$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a $ = $\widehat i - \widehat j$, $\overrightarrow b $ = $\widehat i + \widehat j + \widehat k$ and $\overrightarrow c $
be a vector such that $\overrightarrow a $ × $\overrightarrow c $ + $\overrightarrow b $ = $\overrightarrow 0 $
and $\overrightarrow a $ . $\overrightarrow c $ = 4, then |$\overrightarrow c $|2 is equal to :
A.
8
B.
$19 \over 2$
C.
9
D.
$17 \over 2$
Correct Answer: B
Explanation:
Given that,
$\overrightarrow a \times \overrightarrow c + \overrightarrow b = \overrightarrow 0 $
$ \Rightarrow $ $\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow c } \right) + \overrightarrow a \times \overrightarrow b = \overrightarrow 0 $
$ \Rightarrow $ $\left( {\overrightarrow a \cdot \overrightarrow c } \right)\overrightarrow a - \left( {\overrightarrow a \cdot \overrightarrow a } \right)\overrightarrow c + \overrightarrow a \times \overrightarrow b = \overrightarrow 0 $
given that
$\overrightarrow a \cdot \overrightarrow c = 4$
and $\overrightarrow a \cdot \overrightarrow a = {\left| {\overrightarrow a } \right|^2} = {\left( {\sqrt 2 } \right)^2} = 2$
$ \Rightarrow $ $4\overrightarrow a - 2\overrightarrow c + \overrightarrow a \times \overrightarrow b = 0$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow c = \widehat j - \widehat k$ and a vector $\overrightarrow b $ be such that $\overrightarrow a \times \overrightarrow b = \overrightarrow c $ and $\overrightarrow a .\overrightarrow b = 3.$ Then $\left| {\overrightarrow b } \right|$ equals :
A.
${{11} \over 3}$
B.
${{11} \over {\sqrt 3 }}$
C.
$\sqrt {{{11} \over 3}} $
D.
${{\sqrt {11} } \over 3}$
Correct Answer: C
Explanation:
$ \because $ $\overrightarrow a $ $=$ $\widehat i + \widehat j + \widehat k \Rightarrow \left| {\overrightarrow a } \right| = \sqrt 3 $
& $\overrightarrow c = \widehat j - \widehat k \Rightarrow \left| {\overrightarrow c } \right|\sqrt 2 $
Now, $\overrightarrow a $ $ \times $ $\overrightarrow b $ = $\overrightarrow c $ (Given)
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow u $ be a vector coplanar with the vectors $\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k$ and $\overrightarrow b = \widehat j + \widehat k$. If $\overrightarrow u $ is perpendicular to $\overrightarrow a $ and $\overrightarrow u .\overrightarrow b = 24$, then ${\left| {\overrightarrow u } \right|^2}$ is equal to
A.
336
B.
315
C.
256
D.
84
Correct Answer: A
Explanation:
You should know that, when $\overrightarrow u $ is coplanar with $\overrightarrow a $ and $\overrightarrow b $ then we can write $\overrightarrow u = x\overrightarrow a + y\overrightarrow b $
Here, $\overrightarrow u $ is perpendicular with $\overrightarrow a $ then,
$\overrightarrow u .\overrightarrow a = 0$
$ \Rightarrow \,\,\,\,\left( {x\,\overrightarrow a + y\overrightarrow b } \right)\,.\overrightarrow a = 0$
$ \Rightarrow \,\,\,\,x\,.\overrightarrow {\,a} \,.\,\overrightarrow a \, + \,y\,.\,\overrightarrow a \,.\,\overrightarrow b = 0$
$ \Rightarrow \,\,\,\,x\,.\,{\left| {\overrightarrow a } \right|^2} + \,y\overrightarrow a \,.\,\overrightarrow b = 0$
$ \Rightarrow \,\,\,\,x + y = 12.......\left( 2 \right)$
By solvig (1) and (2) we get,
x = - 2 and y = 14
Now, ${\left| {\overrightarrow u } \right|^2} = \overrightarrow u .\overrightarrow u $
$ = \,\,\,\,\left( {x\overrightarrow a + y\overrightarrow b } \right).\overrightarrow u $
$ = \,\,\,\,x\overrightarrow a .\overrightarrow u + y\overrightarrow b .\overrightarrow u $
$ = \,\,\,\,x\overrightarrow a .\overrightarrow u + y\overrightarrow b .\overrightarrow u $
$ = \,\,\,\,0 + 14 \times 24$ [as $\overrightarrow a .\overrightarrow u = 0$ and $\overrightarrow u .\overrightarrow b = 24]$
$=\,\,\,\,$ 336
2018
Q234
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If the position vectors of the vertices A, B and C of a $\Delta $ ABC are respectively $4\widehat i + 7\widehat j + 8\widehat k,$ $2\widehat i + 3\widehat j + 4\widehat k,$ and $2\widehat i + 5\widehat j + 7\widehat k,$ then the position vectors of the point, where the bisector of $\angle $A meets BC is :
Therefore, position vector of point p = ${1 \over 3}$ (6i + 13j + 18k)
2018
Q235
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If $\overrightarrow a ,\,\,\overrightarrow b ,$ and $\overrightarrow C $ are unit vectors such that $\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 ,$ then $\left| {\overrightarrow a \times \overrightarrow c } \right|$ is equal to :
A.
${{\sqrt {15} } \over 4}$
B.
${{1} \over {4}}$
C.
${{15} \over {16}}$
D.
${{\sqrt {15} } \over 16}$
Correct Answer: A
Explanation:
Given,
$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 $
$ \Rightarrow $ $\overrightarrow a + 2\overrightarrow c = - 2\overrightarrow b $
Squaring both sides,
${\left| {\overrightarrow a } \right|^2} + 4\overrightarrow a .\overrightarrow c + 4{\left| {\overrightarrow c } \right|^2} = 4{\left| {\overrightarrow b } \right|^2}$
$ \Rightarrow $ 1 + $4\overrightarrow a .\overrightarrow c $ + 4 = 4 [as $\left| {\overrightarrow a } \right|^2$ = $\left| {\overrightarrow b } \right|^2$ = $\left| {\overrightarrow c } \right|^2$ = 1]
$ \Rightarrow $ $\overrightarrow a .\overrightarrow c = - {1 \over 4}$
$ \Rightarrow $ $\left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos \theta = - {1 \over 4}$
$ \therefore $ $\left| {\overrightarrow a \times \overrightarrow c } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \theta $
= 1 . 1 . ${{\sqrt {15} } \over 4}$
= ${{\sqrt {15} } \over 4}$
2017
Q236
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If the vector $\overrightarrow b = 3\widehat j + 4\widehat k$ is written as the
sum of a vector $\overrightarrow {{b_1}} ,$ paralel to $\overrightarrow a = \widehat i + \widehat j$ and a vector $\overrightarrow {{b_2}} ,$ perpendicular to $\overrightarrow a ,$ then $\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} $ is equal to :
A.
$ - 3\widehat i + 3\widehat j - 9\widehat k$
B.
$6\widehat i - 6\widehat j + {9 \over 2}\widehat k$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8\widehat i - 6\widehat j$ and $3\widehat i + 4\widehat j - 12\widehat k,$ is :
A.
26
B.
65
C.
20
D.
52
Correct Answer: B
Explanation:
When diagonal ${\overrightarrow {{d_1}} }$ and ${\overrightarrow {{d_2}} }$ are given of a parallelogram then the area of parallelogram = ${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a = 2\widehat i + \widehat j -2 \widehat k$ and $\overrightarrow b = \widehat i + \widehat j$.
Let $\overrightarrow c $ be a vector such that $\left| {\overrightarrow c - \overrightarrow a } \right| = 3$,
$\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3$ and the angle between $\overrightarrow c $ and $\overrightarrow a \times \overrightarrow b$ is $30^\circ $.
Then $\overrightarrow a .\overrightarrow c $ is equal to :
A.
2
B.
5
C.
${1 \over 8}$
D.
${{25} \over 8}$
Correct Answer: A
Explanation:
Given:
$\overrightarrow a = 2\widehat i + \widehat j - 2\widehat k,\,\,\overrightarrow b = \widehat i + \widehat j$
$ \Rightarrow $ $\left| {\overrightarrow a } \right| = 3$
$ \therefore $ $\overrightarrow a \times \overrightarrow b = 2\widehat i - 2\widehat j + \widehat k$
$\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{2^2} + {2^2} + {1^2}} = 3$
We have $\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \left| {\overrightarrow a \times \overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin 30^\circ$
$ \Rightarrow $ $\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3\left| {\overrightarrow c } \right|.{1 \over 2}$
$ \therefore $ $\left| {\overrightarrow c } \right| = 2$
Now $\left| {\overrightarrow c - \overrightarrow a } \right| = 3$
On squaring, we get
$ \Rightarrow $ ${c^2} + {a^2} - 2 - \overrightarrow c .\overrightarrow a = 9$
$ \Rightarrow $ $4 + 9 - 2 - \overrightarrow a .\overrightarrow c = 9$
$ \Rightarrow $ $\overrightarrow a .\overrightarrow c = 2$ [$ \because $ $\overrightarrow c .\overrightarrow a \,\, = \,\,\overrightarrow a .\overrightarrow c $]
2016
Q239
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ and ${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$ respectively, then the position vector of the orthocentre of this triangle, is :
A.
${\overrightarrow a + \overrightarrow b + \overrightarrow c }$
B.
$ - \left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)$
C.
$\overrightarrow 0 $
D.
$\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)$
Correct Answer: D
Explanation:
Given,
Position vector of circumcentre, $\overrightarrow C = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$
We know, position vector of centroid, $\overrightarrow G = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 3}$
Now, let $\overrightarrow R $ be the orthocentre of the triangle.
We know, $\overrightarrow G $ $ = {{2\overrightarrow C + \overrightarrow R } \over 3}$
$ \Rightarrow $ 3$\overrightarrow G $ $ = 2\overrightarrow C + \overrightarrow R $
$ \Rightarrow $ $\overrightarrow R = 3\overrightarrow G - 2\overrightarrow C $
= $\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) - 2\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}} \right)$
= ${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}$
2016
Q240
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3$\widehat i$ + $\widehat j$ $-$ $\widehat k$, $-$$\widehat i$ + 3$\widehat j$ + p$\widehat k$ and 5$\widehat i$ + q$\widehat j$ $-$ 4$\widehat k$, then the point (p, q) lies
on a line :
A.
parallel to x-axis.
B.
parallel to y-axis.
C.
making an acute angle with the positive direction of x-axis.
D.
making an obtuse angle with the positive direction of x-axis.
Correct Answer: C
Explanation:
Given,
$\overrightarrow A = 3\widehat i + \widehat j - \widehat k$
$\overrightarrow B = - \widehat i + 3\widehat j - p\widehat k$
$\overrightarrow C = 5\widehat i + 9\widehat j - 4\widehat k$
$ \therefore $ line makes an angle less than 90o or acute angle with the positive direction of x-axis.
2016
Q241
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a ,\overrightarrow b $ and $\overrightarrow c $ be three unit vectors such that $\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = {{\sqrt 3 } \over 2}\left( {\overrightarrow b + \overrightarrow c } \right).$ If ${\overrightarrow b }$ is not parallel to ${\overrightarrow c },$ then the angle between ${\overrightarrow a }$ and ${\overrightarrow b }$ is:
A.
${{2\pi } \over 3}$
B.
${{5\pi } \over 6}$
C.
${{3\pi } \over 4}$
D.
${{\pi } \over 2}$
Correct Answer: B
Explanation:
$\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = {{\sqrt 3 } \over 2}\left( {\overrightarrow b + \overrightarrow c } \right)$
$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c = {{\sqrt 3 } \over 2}\overrightarrow b + {{\sqrt 3 } \over 2}\overrightarrow c $
$\left[ \, \right.$ As $\overrightarrow a $ and $\overrightarrow b $ are unit vectors $\left. \, \right]$
where $\theta $ is the angle between $\overrightarrow a $ and $\overrightarrow b $
$\theta = {{5\pi } \over 6}$
2015
Q242
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a ,\overrightarrow b $ and $\overrightarrow c $ be three non-zero vectors such that no two of them are collinear and
$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a .$ If $\theta $ is the angle between vectors $\overrightarrow b $ and ${\overrightarrow c }$ , then a value of sin $\theta $ is :
A.
${2 \over 3}$
B.
${{ - 2\sqrt 3 } \over 3}$
C.
${{ 2\sqrt 2 } \over 3}$
D.
${{ - \sqrt 2 } \over 3}$
Correct Answer: C
Explanation:
$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$ \Rightarrow - \overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$ \Rightarrow - \left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a + \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$ \Rightarrow - \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\cos \theta \overrightarrow a + \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$\therefore$ $\,\,\,\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $ are non collinear, the above equation is possible only when
$ - \cos \theta = {1 \over 3}$ and $\overrightarrow c .\overrightarrow a = 0$
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If $\left[ {\overrightarrow a \times \overrightarrow b \,\,\,\,\overrightarrow b \times \overrightarrow c \,\,\,\,\overrightarrow c \times \overrightarrow a } \right] = \lambda {\left[ {\overrightarrow a\,\,\,\,\,\,\,\, \overrightarrow b \,\,\,\,\,\,\,\,\overrightarrow c } \right]^2}$ then $\lambda $ is equal to :
A.
$0$
B.
$1$
C.
$2$
D.
$3$
Correct Answer: B
Explanation:
$L.H.S$ $ = \left( {\overrightarrow a \times \overrightarrow b } \right).\left[ {\left( {\overrightarrow b \times \overrightarrow c } \right) \times \left( {\overrightarrow c \times \overrightarrow a } \right)} \right]$
$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\left[ {\left( {\overrightarrow b \times \overrightarrow c .\overrightarrow a } \right)} \right]\overrightarrow c - \left( {\overrightarrow b \times \overrightarrow c .\overrightarrow c } \right)\left. {\overrightarrow a } \right]$
$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\left[ {\left[ {\overrightarrow b \,\overrightarrow c \,\overrightarrow a } \right]\overrightarrow c } \right]$ $\,\,\,\,\,\,\left[ \, \right.$As $\overrightarrow b \times \overrightarrow c .\overrightarrow c = 0$ $\left. \, \right]$
$ = \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right].\left( {\overrightarrow a \times \overrightarrow b .\overrightarrow c } \right) = {\left[ {\overrightarrow a \,\,\,\,\,\overrightarrow b \,\,\,\,\,\overrightarrow c } \right]^2}$
$\left[ {\overrightarrow a \times \overrightarrow b \,\,\,\overrightarrow b \times \overrightarrow c \,\,\,\overrightarrow c \times \overrightarrow a } \right] = {\left[ {\overrightarrow a \,\,\,\,\,\overrightarrow b \,\,\,\,\,\overrightarrow c } \right]^2}$
So $\,\,\,\,\,\lambda = 1$
2013
Q244
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If the vectors $\overrightarrow {AB} = 3\widehat i + 4\widehat k$ and $\overrightarrow {AC} = 5\widehat i - 2\widehat j + 4\widehat k$ are the sides of a triangle $ABC,$ then the length of the median through $A$ is :
$ = 4\overrightarrow i + \overrightarrow j + 4\overrightarrow k $
Length of median $AM$
$ = \sqrt {16 + 1 + 16} = \sqrt {33} $
2012
Q245
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a $ and $\overrightarrow b $ be two unit vectors. If the vectors $\,\overrightarrow c = \widehat a + 2\widehat b$ and $\overrightarrow d = 5\widehat a - 4\widehat b$ are perpendicular to each other, then the angle between $\overrightarrow a $ and $\overrightarrow b $ is :
A.
${\pi \over 6}$
B.
${\pi \over 2}$
C.
${\pi \over 3}$
D.
${\pi \over 4}$
Correct Answer: C
Explanation:
Let $\overrightarrow c = \widehat a + 2\widehat b$ and $\overrightarrow d = 5\widehat a - 4\widehat b$
Since $\overrightarrow c $ and $\overrightarrow d $ are perpendicular to each other
$\therefore$ $\overrightarrow c .\overrightarrow d = 0 \Rightarrow \left( {\widehat a + 2\widehat b} \right).\left( {5\widehat a - 4\widehat b} \right) = 0$
$ \Rightarrow 5 + 6\widehat a.\widehat b - 8 = 0$ $\,\,\,\,\,\,$ (as $\widehat a.\widehat a = 1$)
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $ABCD$ be a parallelogram such that $\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p $ and $\angle BAD$ be an acute angle. If $\overrightarrow r $ is the vector that coincide with the altitude directed from the vertex $B$ to the side $AD,$ then $\overrightarrow r $ is given by :
A.
$\overrightarrow r = 3\overrightarrow q - {{3\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}\overrightarrow p $
B.
$\overrightarrow r = - \overrightarrow q + {{\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}\overrightarrow p $
$\overrightarrow r = - 3\overrightarrow q - {{3\left( {\overrightarrow p .\overrightarrow q } \right)} \over {\left( {\overrightarrow p .\overrightarrow p } \right)}}$
Correct Answer: B
Explanation:
Let $ABCD$ be a parallelogram such that
$\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p $ and $\angle BAD$ be an acute angle.
We have
$\overrightarrow {AX} = \left( {{{\overrightarrow p .\overrightarrow q } \over {\left| {\overrightarrow p } \right|}}} \right)\left( {{{\overrightarrow p } \over {\left| {\overrightarrow p } \right|}}} \right) = {{\overrightarrow p .\overrightarrow q } \over {{{\left| {\overrightarrow p } \right|}^2}}}\overrightarrow p $
Let $\overrightarrow r = \overrightarrow {BX} = \overrightarrow {BA} + \overrightarrow {AX} $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \overrightarrow q + {{\overrightarrow p .\overrightarrow q } \over {{{\left| {\overrightarrow p } \right|}^2}}}\overrightarrow p $
2011
Q247
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The vectors $\overrightarrow a $ and $\overrightarrow b $ are not perpendicular and $\overrightarrow c $ and $\overrightarrow d $ are two vectors satisfying $\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow d $ and $\overrightarrow a .\overrightarrow d = 0\,\,.$ Then the vector $\overrightarrow d $ is equal to :
A.
$\overrightarrow c + \left( {{{\overrightarrow a .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow b $
B.
$\overrightarrow b + \left( {{{\overrightarrow b .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow c $
C.
$\overrightarrow c - \left( {{{\overrightarrow a .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow b $
D.
$\overrightarrow b - \left( {{{\overrightarrow b .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow c $
Correct Answer: C
Explanation:
$\overrightarrow a .\overrightarrow b \ne 0,\overrightarrow a .\overrightarrow d = 0$
Now, $\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow d $
$ \Rightarrow \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow d } \right)$
$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c = \left( {\overrightarrow a .\overrightarrow d } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow d $
$ \Rightarrow \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow d = - \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b + \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c $
$\overrightarrow d = \overrightarrow c - \left( {{{\overrightarrow a .\overrightarrow c } \over {\overrightarrow a .\overrightarrow b }}} \right)\overrightarrow b $
2011
Q248
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If $\overrightarrow a = {1 \over {\sqrt {10} }}\left( {3\widehat i + \widehat k} \right)$ and $\overrightarrow b = {1 \over 7}\left( {2\widehat i + 3\widehat j - 6\widehat k} \right),$ then the value
of $\left( {2\overrightarrow a - \overrightarrow b } \right)\left[ {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a + 2\overrightarrow b } \right)} \right]$ is :
A.
$-3$
B.
$5$
C.
$3$
D.
$-5$
Correct Answer: D
Explanation:
We have $\overrightarrow a .\overrightarrow b = 0,\,\,\overrightarrow a .\overrightarrow a = 1,\,\,\overrightarrow b .\overrightarrow b = 1$
$\left( {2\overrightarrow a - \overrightarrow b } \right).\left[ {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a + 2\overrightarrow b } \right)} \right]$
$ = \left( {2\overrightarrow a - \overrightarrow b } \right).\left[ {\left\{ {\overrightarrow a .\left( {\overrightarrow a + 2\overrightarrow b } \right)} \right\}\overrightarrow b - \left\{ {\overrightarrow b .\left( {\overrightarrow a + 2\overrightarrow b } \right)\overrightarrow a } \right\}} \right]$
$ = \left( {2\overrightarrow a - \overrightarrow b } \right).\left[ {\left( {\overrightarrow a .\overrightarrow a + 2\overrightarrow a .\overrightarrow b } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b + 2\overrightarrow b .\overrightarrow b } \right)\overrightarrow a } \right]$
$ = \left( {2\overrightarrow a - \overrightarrow b } \right).\left[ {\overrightarrow b - 2\overrightarrow a } \right]$
$ = 4\overrightarrow a .\overrightarrow b - \overrightarrow b .\overrightarrow b - 4\overrightarrow a .\overrightarrow a $
$ = - 5$
2011
Q249
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Let $\overrightarrow a $, $\overrightarrow b $, $\overrightarrow c $ be three non-zero vectors which are pairwise non-collinear. If $\overrightarrow a+3 \overrightarrow b$ is collinear with $\overrightarrow c$ and $\overrightarrow b+2 \overrightarrow c$ is collinear with $\overrightarrow a$, then $\overrightarrow a+\overrightarrow b+6 \overrightarrow c$ is :
A.
$\overrightarrow a+\overrightarrow c$
B.
$\overrightarrow c$
C.
$\overrightarrow a$
D.
$\overrightarrow 0$
Correct Answer: D
Explanation:
We are given that $\overrightarrow a + 3 \overrightarrow b$ is collinear with $\overrightarrow c$, and $\overrightarrow b + 2 \overrightarrow c$ is collinear with $\overrightarrow a$. This means we can write:
$\overrightarrow a + 3 \overrightarrow b = \lambda \overrightarrow c \quad ...(i)$
$\overrightarrow b + 2 \overrightarrow c = \mu \overrightarrow a \quad ...(ii)$
for some scalars $\lambda$ and $\mu$.
We are trying to find $\overrightarrow a + \overrightarrow b + 6\overrightarrow c$ in terms of $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$. We can also express this as :
$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = (\lambda + 6) \overrightarrow c \quad ...(iii)$
by adding $6\overrightarrow c$ to both sides of equation (i).
Now, from equation (ii), multiplying by 3 gives us :
$3\overrightarrow b + 6 \overrightarrow c = 3\mu \overrightarrow a \quad ...(iv)$
Adding $\overrightarrow a$ to both sides of equation (iv) gives :
$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = (1 + 3\mu) \overrightarrow a \quad ...(v)$
Now, we have two expressions for $\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c$, one in terms of $\overrightarrow c$ (from equation iii) and one in terms of $\overrightarrow a$ (from equation v). Setting these equal to each other gives :
$(\lambda + 6) \overrightarrow c = (1 + 3\mu) \overrightarrow a \quad ...(vi)$
Since $\overrightarrow a$ and $\overrightarrow c$ are not collinear, this equation can only hold if the coefficients on both sides are zero, hence :
$\lambda + 6 = 0$ and $1 + 3\mu = 0$
This gives $\lambda = -6$ and $\mu = -\frac{1}{3}$.
Finally, substituting $\lambda = -6$ into equation (iii) gives :
$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = 0$
So, $\overrightarrow a + \overrightarrow b + 6\overrightarrow c = \overrightarrow 0$.
Therefore, the correct answer is Option D : $\overrightarrow 0$.
2010
Q250
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
If the vectors $\overrightarrow a = \widehat i - \widehat j + 2\widehat k,\,\,\,\,\,\overrightarrow b = 2\widehat i + 4\widehat j + \widehat k\,\,\,$ and $\,\overrightarrow c = \lambda \widehat i + \widehat j + \mu \widehat k$ are mutually orthogonal, then $\,\left( {\lambda ,\mu } \right)$ is equal to :
A.
$(2, -3)$
B.
$(-2, 3)$
C.
$(3, -2)$
D.
$(-3, 2)$
Correct Answer: D
Explanation:
Since, $\overrightarrow a ,\overrightarrow b $ and $\overrightarrow c $ are mutually orthogonal
$\overrightarrow a .\overrightarrow b = 0,\,\,\overrightarrow b .\overrightarrow c = 0,\,\,\overrightarrow c .\overrightarrow a = 0$
