Vector Algebra
619 Questions
1988
JEE Advanced
Numerical
IIT-JEE 1988
Let $OA$ $CB$ be a parallelogram with $O$ at the origin and $OC$ a diagonal. Let $D$ be the midpoint of $OA.$ Using vector methods prove that $BD$ and $CO$ intersect in the same ratio. Determine this ratio.
Correct Answer: $$2:1$$
1988
JEE Advanced
Numerical
IIT-JEE 1988
The components of a vector $\overrightarrow a $ along and perpendicular to a non-zero vector $\overrightarrow b $ are ......and .....respectively.
Correct Answer: $$\,\left( {{{\overrightarrow {a\,} .\,\overrightarrow b } \over {{{\left| {\overrightarrow b } \right|}^2}}}} \right)\overrightarrow b \,.\,,\,\,\overrightarrow a - \left( {{{\overrightarrow a \,.\,\overrightarrow b } \over {\left| {\overrightarrow b } \right|}}} \right)\overrightarrow b $$
1987
JEE Advanced
MCQ
IIT-JEE 1987
The number of vectors of unit length perpendicular to vectors $\overrightarrow a = \left( {1,1,0} \right)$ and $\overrightarrow b = \left( {0,1,1} \right)$ is
A.
one
B.
two
C.
three
D.
infinite
1987
JEE Advanced
Numerical
IIT-JEE 1987
If $A, B, C, D$ are any four points in space, prove that -
$\left| {\overrightarrow {AB} \times \overrightarrow {CD} + \overrightarrow {BC} \times \overrightarrow {AD} + \overrightarrow {CA} \times \overrightarrow {BD} } \right| = 4$ (area of triangle $ABC$)
$\left| {\overrightarrow {AB} \times \overrightarrow {CD} + \overrightarrow {BC} \times \overrightarrow {AD} + \overrightarrow {CA} \times \overrightarrow {BD} } \right| = 4$ (area of triangle $ABC$)
Correct Answer: Solve it.
1987
JEE Advanced
Numerical
IIT-JEE 1987
If the vectors $a\widehat i + \widehat j + \widehat k,\,\,\widehat i + b\widehat j + \widehat k$ and $\widehat i + \widehat j + c\widehat k$
$\left( {a \ne b \ne c \ne 1} \right)$ are coplannar, then the value of ${1 \over {\left( {1 - a} \right)}} + {1 \over {\left( {1 - b} \right)}} + {1 \over {\left( {1 - c} \right)}} = ..........$
$\left( {a \ne b \ne c \ne 1} \right)$ are coplannar, then the value of ${1 \over {\left( {1 - a} \right)}} + {1 \over {\left( {1 - b} \right)}} + {1 \over {\left( {1 - c} \right)}} = ..........$
Correct Answer: $$1$$
1987
JEE Advanced
Numerical
IIT-JEE 1987
Let $b = 4\widehat i + 3\widehat j$ and $\overrightarrow c $ be two vectors perpendicular to each other in the $xy$-plane. All vectors in the same plane having projecttions $1$ and $2$ along $\overrightarrow b $ and $\overrightarrow c, $ respectively, are given by ...........
Correct Answer: $$2\widehat i - \widehat j$$
1986
JEE Advanced
MCQ
IIT-JEE 1986
Let $\overrightarrow a = {a_1}i + {a_2}j + {a_3}k,\,\,\,\overrightarrow b = {b_1}i + {b_2}j + {b_3}k$ and $\overrightarrow c = {c_1}i + {c_2}j + {c_3}k$ be three non-zero vectors such that $\overrightarrow c $ is a unit vector perpendicular to both the vectors $\overrightarrow a $ and $\overrightarrow b .$ If the angle between $\overrightarrow a $ and $\overrightarrow b $ is ${\pi \over 6},$ then
${\left| {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{b_1}} & {{b_2}} & {{b_3}} \cr {{c_1}} & {{c_2}} & {{c_3}} \cr } } \right|^2}$ is equal to
${\left| {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{b_1}} & {{b_2}} & {{b_3}} \cr {{c_1}} & {{c_2}} & {{c_3}} \cr } } \right|^2}$ is equal to
A.
$0$
B.
$1$
C.
${1 \over 4}\left( {a_1^2 + a_2^2 + a_2^3} \right)\left( {b_1^2 + b_2^2 + b_3^2} \right)$
D.
${3 \over 4}\left( {a_1^2 + a_2^2 + a_3^2} \right)\left( {b_1^2 + b_2^2 + b_3^2} \right)\left( {c_1^2 + c_2^2 + c_3^2} \right)$
1986
JEE Advanced
Numerical
IIT-JEE 1986
The position vectors of the points $A, B, C$ and $D$ are $3\widehat i - 2\widehat j - \widehat k,\,2\widehat i + 3\widehat j - 4\widehat k,\, - \widehat i + \widehat j + 2\widehat k$ and $4\widehat i + 5\widehat j + \lambda \widehat k,$
respectively. If the points $A, B, C$ and $D$ lie on a plane, find the value of $\lambda .$
respectively. If the points $A, B, C$ and $D$ lie on a plane, find the value of $\lambda .$
Correct Answer: $${{146} \over {17}}$$
1985
JEE Advanced
Numerical
IIT-JEE 1985
If $\overrightarrow A = \left( {1,1,1} \right),\,\,\overrightarrow C = \left( {0,1, - 1} \right)$ are given vectors, then a vector $B$ satifying the equations $\overrightarrow A \times \overrightarrow B = \overrightarrow {\,C} $ and $\overrightarrow A .\overrightarrow B = \overrightarrow {3\,} $ ..........
Correct Answer: $${5 \over 3}\widehat i + {2 \over 3}\widehat j + {2 \over 3}\widehat k$$
1985
JEE Advanced
Numerical
IIT-JEE 1985
If $\overrightarrow A \overrightarrow {\,B} \overrightarrow {\,C} $ are three non-coplannar vectors, then -
${{\overrightarrow A .\overrightarrow B \times \overrightarrow C } \over {\overrightarrow C \times \overrightarrow A .\overrightarrow B }} + {{\overrightarrow B .\overrightarrow A \times \overrightarrow C } \over {\overrightarrow C .\overrightarrow A \times \overrightarrow B }} = $ ................
${{\overrightarrow A .\overrightarrow B \times \overrightarrow C } \over {\overrightarrow C \times \overrightarrow A .\overrightarrow B }} + {{\overrightarrow B .\overrightarrow A \times \overrightarrow C } \over {\overrightarrow C .\overrightarrow A \times \overrightarrow B }} = $ ................
Correct Answer: $$0$$
1984
JEE Advanced
Numerical
IIT-JEE 1984
$A, B, C$ and $D,$ are four points in a plane with position vectors $a, b, c$ and $d$ respectively such that
$$\left( {\overrightarrow a - \overrightarrow d } \right)\left( {\overrightarrow b - \overrightarrow c } \right) = \left( {\overrightarrow b - \overrightarrow d } \right)\left( {\overrightarrow c - \overrightarrow a } \right) = 0$$
The point $D,$ then, is the ................ of the triangle $ABC.$
Correct Answer: Orthocenter
1984
JEE Advanced
MCQ
IIT-JEE 1984
The points with position vectors $a+b,$ $a-b,$ and $a+kb$ are collinear for all real values of $k.$
A.
TRUE
B.
FALSE
1983
JEE Advanced
MCQ
IIT-JEE 1983
If $X.A=0, X.B=0, X.C=0$ for some non-zero vector $X,$ then $\left[ {A\,B\,C} \right] = 0$
A.
TRUE
B.
FALSE
1982
JEE Advanced
MCQ
IIT-JEE 1982
For non-zero vectors ${\overrightarrow a ,\,\overrightarrow b ,\overrightarrow c },$ $\left| {\left( {\overrightarrow a \times \overrightarrow b } \right).\overrightarrow c } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|$ holds if and only if
A.
$\overrightarrow a \,.\,\overrightarrow b = 0,\overrightarrow b \,.\,\overrightarrow c = 0$
B.
$\overrightarrow b \,.\,\overrightarrow c = 0,\overrightarrow c \,.\,\overrightarrow a = 0$
C.
$\overrightarrow c \,.\,\overrightarrow a = 0,\overrightarrow a \,.\,\overrightarrow b = 0$
D.
$\overrightarrow a \,.\,\overrightarrow b = \overrightarrow b \,.\,\overrightarrow c = \overrightarrow c \,.\,\overrightarrow a = 0$
1982
JEE Advanced
Numerical
IIT-JEE 1982
Find all values of $\lambda $ such that $x, y, z,$$\, \ne $$(0,0,0)$ and
$\left( {\overrightarrow i + \overrightarrow j + 3\overrightarrow k } \right)x + \left( {3\overrightarrow i - 3\overrightarrow j + \overrightarrow k } \right)y + \left( { - 4\overrightarrow i + 5\overrightarrow j } \right)z$
$ = \lambda \left( {x\overrightarrow i \times \overrightarrow j \,\,y + \overrightarrow k \,z} \right)$ where $\overrightarrow i ,\,\,\overrightarrow j ,\,\,\overrightarrow k $ are unit vectors along the coordinate axes.
$\left( {\overrightarrow i + \overrightarrow j + 3\overrightarrow k } \right)x + \left( {3\overrightarrow i - 3\overrightarrow j + \overrightarrow k } \right)y + \left( { - 4\overrightarrow i + 5\overrightarrow j } \right)z$
$ = \lambda \left( {x\overrightarrow i \times \overrightarrow j \,\,y + \overrightarrow k \,z} \right)$ where $\overrightarrow i ,\,\,\overrightarrow j ,\,\,\overrightarrow k $ are unit vectors along the coordinate axes.
Correct Answer: $$\lambda = 0,\, - 1$$
1982
JEE Advanced
Numerical
IIT-JEE 1982
${A_1},{A_2},.................{A_n}$ are the vertices of a regular plane polygon with $n$ sides and $O$ is its centre. Show that
$\sum\limits_{i = 1}^{n - 1} {\left( {\overrightarrow {O{A_i}} \times {{\overrightarrow {OA} }_{i + 1}}} \right) = \left( {1 - n} \right)\left( {{{\overrightarrow {OA} }_2} \times {{\overrightarrow {OA} }_1}} \right)} $
$\sum\limits_{i = 1}^{n - 1} {\left( {\overrightarrow {O{A_i}} \times {{\overrightarrow {OA} }_{i + 1}}} \right) = \left( {1 - n} \right)\left( {{{\overrightarrow {OA} }_2} \times {{\overrightarrow {OA} }_1}} \right)} $
Correct Answer: Solve it
1981
JEE Advanced
MCQ
IIT-JEE 1981
The scalar $\overrightarrow A .\left( {\overrightarrow B + \overrightarrow C } \right) \times \left( {\overrightarrow A + \overrightarrow B + \overrightarrow C } \right)$ equals :
A.
$0$
B.
$\left[ {\overrightarrow A \,\overrightarrow B \,\overrightarrow C } \right] + \left[ {\overrightarrow B \,\overrightarrow C \,\overrightarrow A } \right]$
C.
$\left[ {\overrightarrow A \,\overrightarrow B \,\overrightarrow C } \right]$
D.
None of these
1981
JEE Advanced
Numerical
IIT-JEE 1981
Let $\overrightarrow A ,\overrightarrow B ,\overrightarrow C $ be vectors of length $3, 4, 5$ respectively. Let $\overrightarrow A $ be perpendicular to $\overrightarrow B + \overrightarrow C ,\overrightarrow B $ to $\overrightarrow C + \overrightarrow A $ to $\overrightarrow A + \overrightarrow B .$ Then the length of vector $\overrightarrow A + \overrightarrow B + \overrightarrow C $ is ..........
Correct Answer: $$5\sqrt 2 $$
1981
JEE Advanced
MCQ
IIT-JEE 1981
Let $\overrightarrow A ,\overrightarrow B $ and ${\overrightarrow C }$ be unit vectors suppose that $\overrightarrow A .\overrightarrow B = \overrightarrow A .\overrightarrow C = 0,$ and thatthe angle between ${\overrightarrow B }$ and ${\overrightarrow C }$ is $\pi /6.$ Then $\overrightarrow A = \pm 2\left( {\overrightarrow B \times \overrightarrow C } \right).$
A.
TRUE
B.
FALSE