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JEE • MATHEMATICS • ADDITION OF VECTORS

Addition of vectors practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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31 Medium
31 of 49
Application +4 | -1
If the points whose position, vectors are 3i−2j−k3 \mathbf{i}-2 \mathbf{j}-\mathbf{k}, 2i+3j−4k,−i+j+2k2 i+3 \mathbf{j}-4 \mathbf{k},-\mathbf{i}+\mathbf{j}+2 \mathbf{k} and 4i+5j+λk4 \mathbf{i}+5 \mathbf{j}+\lambda \mathbf{k} lie on a plane, then λ=\lambda=
View answer and explanation
Correct option: A
−14617-\frac{146}{17}
Option A (-146/17) is correct because the points are coplanar, so the scalar triple product of the vectors formed by these points must satisfy the coplanarity condition.
32 Medium
32 of 49
Application +4 | -1
The position vector of the point in which the line joining the points i−2j+k\mathbf{i}-2 \mathbf{j}+\mathbf{k} and 3k−2j3 \mathbf{k}-2 \mathbf{j} cuts the plane through the origin and the points 4j4 \mathbf{j} and 2i+k2 \mathbf{i}+\mathbf{k}, is
View answer and explanation
Correct option: B
15(6i−10j+3k)\frac{1}{5}(6 \mathbf{i}-10 \mathbf{j}+3 \mathbf{k})
Option B is correct because it represents the intersection point of the line joining the two given points and the plane passing through the origin and the other two points.
33 Medium
33 of 49
Application +4 | -1
The line through i+3j+2k\mathbf{i}+3 \mathbf{j}+2 \mathbf{k} and perpendicular to the lines r=(i+2j−k)+λ(2i+j+k)\mathbf{r}=(\mathbf{i}+2 \mathbf{j}-\mathbf{k})+\lambda(2 \mathbf{i}+\mathbf{j}+\mathbf{k})
View answer and explanation
Correct option: D
r=i+3j+2k+λ(−i+5j−3k)\mathbf{r}=\mathbf{i}+3 \mathbf{j}+2 \mathbf{k}+\lambda(-\mathbf{i}+5 \mathbf{j}-3 \mathbf{k})
Option D is correct because the line passes through the given point and is perpendicular to the two given lines, hence parallel to their cross product.
34 Medium
34 of 49
Analyzing +4 | -1
The unit vectors orthogonal to the vector −i^+2j^+2k^- \hat i + 2\hat j + 2\hat k and making equal angles with the XX and YY axes is (are)
View answer and explanation
Correct option: A
±13(2i^+2j^−k^)\pm \frac{1}{3}\left( {2\hat i + 2\hat j - \hat k} \right)
Option A (±13(2i^+2j^−k^)\pm \frac{1}{3}\left( {2\hat i + 2\hat j - \hat k} \right)) is correct because it satisfies the orthogonality condition and equal angles with the X and Y axes.
35 Medium
35 of 49
Application +4 | -1
Vector coplanar with vectors i+j\mathbf{i}+\mathbf{j} and j+k\mathbf{j}+\mathbf{k} and parallel to the vector 2i−2j−4k2 \mathbf{i}-2 \mathbf{j}-4 \mathbf{k}, is
View answer and explanation
Correct option: B
i−j−2k\mathbf{i}-\mathbf{j}-2 \mathbf{k}
Option B (i−j−2k\mathbf{i}-\mathbf{j}-2 \mathbf{k}) is correct because it satisfies the conditions of being coplanar with i+j\mathbf{i}+\mathbf{j} and j+k\mathbf{j}+\mathbf{k} and parallel to 2i−2j−4k2 \mathbf{i}-2 \mathbf{j}-4 \mathbf{k}.
36 Medium
36 of 49
Application +4 | -1
If the position vectors of the point A,B,CA, B, C be i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} respectively and PP be a point such that AB→=CP→\overrightarrow{A B}=\overrightarrow{C P}, then the position vector of PP is
View answer and explanation
Correct option: A
−i+j+k-\mathbf{i}+\mathbf{j}+\mathbf{k}
Option A (−i+j+k-\mathbf{i}+\mathbf{j}+\mathbf{k}) is correct because it satisfies the condition AB→=CP→\overrightarrow{A B}=\overrightarrow{C P} by finding the position vector of point P.
37 Medium
37 of 49
Application +4 | -1
The position vectors of the points A,B,CA, B, C and DD are 3i−2j−k,2i+3j−4k,−i+j+2k3 {i}-2 {j}-{k}, 2 {i}+3 {j}-4 {k},-{i}+{j}+2 {k} and 4i+5j+λk4 {i}+5 {j}+\lambda {k} respectively. If the points A,B,C{A}, {B}, {C} and D{D} are on a plane, then the value of λ\lambda is
View answer and explanation
Correct option: B
−14617\frac{-146}{17}
Option B (−14617\frac{{-146}}{{17}}) is correct because the points are coplanar if the scalar triple product of the vectors formed by the points is zero, leading to the given value of λ\lambda.
38 Medium
38 of 49
Application +4 | -1
Let two non-collinear unit vectors a^{\mathbf{\hat a}} and b^{\mathbf{\hat b}} from and acute angle. A point PP moves so that at any time tt the position vector OP→\overrightarrow {\mathbf{OP}} (where OO is the origin) is given by a^cos⁡t+b^sin⁡t\mathbf{\hat a}\cos t + \mathbf{\hat b}\sin t. When PP is farthest form origin OO, let MM be the length of OP→\overrightarrow {\mathbf{OP}} and u^{\mathbf{\hat u}} be the unit vector along OP→\overrightarrow {\mathbf{OP}} Then,
View answer and explanation
Correct option: A
u^=a^+b^∣a^+b^∣and  M=(1+a^⋅b^)1/2\mathbf{\hat u} = \frac{{\mathbf{\hat a} + \mathbf{\hat b}}}{{\left| {\mathbf{\hat a} + \mathbf{\hat b}} \right|}}\mathrm{and\;}M = {(1 + \mathbf{\hat a}\cdot\mathbf{\hat b})^{1/2}}
Option A (u^=a^+b^∣a^+b^∣and  M=(1+a^⋅b^)1/2\mathbf{\hat u} = \frac{{\mathbf{\hat a} + \mathbf{\hat b}}}{{\left| {\mathbf{\hat a} + \mathbf{\hat b}} \right|}}\mathrm{and\;}M = {(1 + \mathbf{\hat a}\cdot\mathbf{\hat b})^{1/2}}) is correct because it correctly finds the unit vector and magnitude when P is farthest from the origin.
39 Medium
39 of 49
Application +4 | -1
A vector of length 3 perpendicular to each of the vectors 3i+j−4k3 i+j-4 k and 6i+5j−2k6 \mathbf{i}+5 \mathbf{j}-2 \mathbf{k} is
View answer and explanation
Correct option: A
2i−2j+k2 \mathbf{i}-2 \mathbf{j}+\mathbf{k}
Option A (2i−2j+k2 \mathbf{i}-2 \mathbf{j}+\mathbf{k}) is correct because it is perpendicular to both given vectors and has a length of 3.
40 Medium
40 of 49
Application +4 | -1
If the vectors i+3j−2k,2i−j+4k\mathbf{i}+3 \mathbf{j}-2 \mathbf{k}, 2 \mathbf{i}-\mathbf{j}+4 \mathbf{k} and 3i+2j+xk3 \mathbf{i}+2 \mathbf{j}+x \mathbf{k} are coplanar, then the value of xx is
View answer and explanation
Correct option: B
2
Option B (2) is correct because for the vectors to be coplanar, the determinant of the matrix formed by their components must be zero, leading to x = 2.
41 Medium
41 of 49
Understanding +4 | -1
If a\mathbf{a} and b\mathbf{b} are two non-collinear vectors and xa+yb=0x \mathbf{a}+y \mathbf{b}=0
View answer and explanation
Correct option: C
x=0,y=0x=0, y=0
Option C (x=0,y=0x=0, y=0) is correct because if two non-collinear vectors are linearly combined to produce a zero vector, both scalar coefficients must be zero.
42 Medium
42 of 49
Application +4 | -1
If two points are P(7,−5,11)P(7,-5,11) and Q(−2,8,13)Q(-2,8,13), then the projection of PQ on a straight line with direction cosines 13,23,23\frac{1}{3}, \frac{2}{3}, \frac{2}{3} is
View answer and explanation
Correct option: A
7
Option A (7) is correct because the projection is calculated using the direction cosines and the difference in coordinates of the two points.
43 Medium
43 of 49
Understanding +4 | -1
The unit vector bisecting OY and OZ is
View answer and explanation
Correct option: C
j^+k^2\frac{{\widehat j + \widehat k}}{{\sqrt 2 }}
Option C (j^+k^2\frac{{\widehat j + \widehat k}}{{\sqrt 2 }}) is correct because it is a unit vector with equal components along the OY and OZ axes.
44 Medium
44 of 49
Application +4 | -1
If ABCDA B C D is a parallelogram and the position vectors of A,BA, B and CC are i^+3j^+5k^,i^+j^+k^\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} and 7i^+7j^+7k^7 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}, then the position vector of DD will be
View answer and explanation
Correct option: B
7i^+9j^+11k^7 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}
Option B is correct because in a parallelogram, the midpoints of the diagonals coincide, so A+C2=B+D2\frac{\mathbf{A} + \mathbf{C}}{2} = \frac{\mathbf{B} + \mathbf{D}}{2}, which leads to D=A+C−B=7i^+9j^+11k^\mathbf{D} = \mathbf{A} + \mathbf{C} - \mathbf{B} = 7 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}.
45 Medium
45 of 49
Application +4 | -1
If the four points, whose position vectors are 3i^−4j^+2k^,i^+2j^−k^,−2i^−j^+3k^3 \hat{i}-4 \hat{j}+2 \hat{k}, \hat{i}+2 \hat{j}-\hat{k},-2 \hat{i}-\hat{j}+3 \hat{k} and 5i^−2αj^+4k^5 \hat{i}-2 \alpha \hat{j}+4 \hat{k} are coplanar, then α\alpha is equal to
View answer and explanation
Correct option: C
7317\frac{73}{17}
Option C (7317\frac{{73}}{{17}}) is correct because it is the value of α\alpha that makes the four points coplanar, calculated using the determinant condition for coplanarity.
46 Medium
46 of 49
Application +4 | -1
A unit vector is orthogonal to 5i^+2j^+6k^5 \hat{i}+2 \hat{j}+6 \hat{k} and is coplanar to 2i^+j^+k^2 \hat{i}+\hat{j}+\hat{k} and i^−j^+k^\hat{i}-\hat{j}+\hat{k} then the vector is
View answer and explanation
Correct option: A
3j^−k^10\frac{3 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{10}}
Option A (3j^−k^10\frac{3 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{10}}) is correct. By setting up a linear combination of the coplanar vectors and using the orthogonality condition, we can solve for the unit vector.
47 Medium
47 of 49
Application +4 | -1
Let p⃗\vec{p} and q⃗\vec{q} be the position vectors of PP and QQ respectively, with respect to OO and ∣p⃗∣=p,∣q⃗∣=q|\vec{p}|=p,|\vec{q}|=q. The points R{R} and SS divide PQ{PQ} internally and externally in the ratio 2:32: 3 respectively. If OR→\overrightarrow{O R} and OS→\overrightarrow{O S} are perpendicular, then
View answer and explanation
Correct option: A
9p2=4q29 p^{2}=4 q^{2}
Option A (9 p2=4 q2) is correct because it is derived from the condition that the vectors OR and OS are perpendicular, leading to a dot product of zero.
48 Medium
48 of 49
Application +4 | -1
The position vectors of the points AA and BB with respect to OO are 2i^+2j^+k^2 \hat{{i}}+2 \hat{{j}}+\hat{{k}} and 2i^+4j^+4k^2 \hat{{i}}+4 \hat{{j}}+4 \hat{{k}}. The length of the internal bisector of ∠BOA\angle B O A of △AOB\triangle A O B is
View answer and explanation
Correct option: B
1363\frac{\sqrt{136}}{3}
Option B (1363\frac{{\sqrt{{136}}}}{{3}}) is correct because it is calculated using the angle bisector theorem and vector properties.
49 Medium
49 of 49
Application +4 | -1
aˉ=−iˉ+jˉ,bˉ=jˉ−kˉ,cˉ=iˉ+jˉ+kˉ\bar{a}=-\bar{i}+\bar{j}, \bar{b}=\bar{j}-\bar{k}, \bar{c}=\bar{i}+\bar{j}+\bar{k} then the unit vector in the opposite direction of aˉ+2bˉ−3cˉ\bar{a}+2 \bar{b}-3 \bar{c}
View answer and explanation
Correct option: C
4iˉ+5kˉ41\frac{4 \bar{i}+5 \bar{k}}{\sqrt{41}}
Option C (4iˉ+5kˉ41\frac{{4 \bar{{i}}+5 \bar{{k}}}}{{\sqrt{{41}}}}) is correct because it represents the unit vector in the opposite direction of the calculated vector aˉ+2bˉ−3cˉ\bar{{a}}+2 \bar{{b}}-3 \bar{{c}}.
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