If the position vector of $A$ with respect to $O$ is $3\widehat {\bf{i}} - 2\widehat {\bf{j}} + 4{\bf{\hat k}}$ and $\overrightarrow {{\bf{AB}}} = 3\widehat {\bf{i}} - \widehat {\bf{j}} + {\bf{\hat k}}$ Then the position vector of $B$ with respect to $O$ is

$6\widehat {\bf{i}} - 3\widehat {\bf{j}} + 5{\bf{\hat k}}$

$ - \widehat {\bf{j}} + 3{\bf{\hat k}}$

$\widehat {\bf{j}} - 3{\bf{\hat k}}$

$\widehat {\bf{i}} - 3\widehat {\bf{j}} + 5{\bf{\hat k}}$

If the vectors ${\bf{\vec a}} = 2\widehat {\bf{i}} + 3\widehat {\bf{j}} + 6{\bf{\hat k}}$ and ${{\bf{\vec b}}}$ are collinear and $\left| {{\bf{\vec b}}} \right| = 21,\;{\rm{then\;}}{\bf{\vec b}}$ is equal to

$ \pm 3\left( {2\widehat {\bf{i}} + 3\widehat {\bf{j}} + 6{\bf{\hat k}}} \right)$

$ \pm \left( {2\widehat {\bf{i}} + 3\widehat {\bf{j}} + 6{\bf{\hat k}}} \right)$

$\left( {\widehat {\bf{i}} + \widehat {\bf{j}} + {\bf{\hat k}}} \right)$

$ \pm 21\left( {2\widehat {\bf{i}} + 3\widehat {\bf{j}} + 6{\bf{\hat k}}} \right)$

If $\widehat {\bf{i}} + 2\widehat {\bf{j}} + 3{\bf{\hat k}}$ and $2\widehat {\bf{i}} - \widehat {\bf{j}} + 4{\bf{\hat k}}$ are the position vectors of the points A and B , then the position vector of the points of trisection of AB are

$\frac{4}{3}\widehat {\bf{i}} + \widehat {\bf{j}} + \frac{{10}}{3}{\bf{\hat k}},\frac{5}{3}\widehat {\bf{i}} + \frac{{11}}{3}{\bf{\hat k}}$

$ - \frac{4}{3}\widehat {\bf{i}} - \widehat {\bf{j}} - \frac{{10}}{3}{\bf{\hat k}},\; - \frac{5}{3}\widehat {\bf{i}} - \frac{{11}}{3}{\bf{\hat k}}$

$\frac{4}{3}\widehat {\bf{i}} - \widehat {\bf{j}} - \frac{{10}}{3}{\bf{\hat k}}, - \frac{5}{3}\widehat {\bf{i}} - \frac{{11}}{3}{\bf{\hat k}}$

$ - \frac{4}{3}\widehat {\bf{i}} + \widehat {\bf{j}} - \frac{{10}}{3}{\bf{\hat k}},\frac{5}{3}\widehat {\bf{i}} - \frac{{11}}{3}{\bf{\hat k}}$

Back to ADDITION OF VECTORS

Your Progress

0 / 318 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • ADDITION OF VECTORS

Addition of vectors practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

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.

Take a chapter quiz →

Showing 271-300 of 318 questions

Preparing practice and saved progress…

271 Medium

271 of 318

Application +4 | -1

If the vectors

\mathbf{i}+3 \mathbf{j}-2 \mathbf{k}, 2 \mathbf{i}-\mathbf{j}+4 \mathbf{k}

and

3 \mathbf{i}+2 \mathbf{j}+x \mathbf{k}

are coplanar, then the value of

x

View answer and explanation

Correct option: B

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.

272 Easy

272 of 318

Application +4 | -1

Let

\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}

and

\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{k}}

, then the point of intersection of the lines

\mathbf{r} \times \mathbf{a}=\mathbf{b} \times \mathbf{a}

and

\mathbf{r} \times \mathbf{b}=\mathbf{a} \times \mathbf{b}

View answer and explanation

Correct option: B

(3,1,-1)

Option B ((3,1,-1)) is correct because it is the point of intersection of the two lines, found by solving the equations derived from the given vector equations.

273 Easy

273 of 318

Application +4 | -1

Three points whose position vectors are

\vec{a}+\vec{b}, \vec{a}-\vec{b}

and

\vec{a}+k \vec{b}

will be collinear, if the value of

k

View answer and explanation

Correct option: D

Every real number

The correct answer is D (Every real number) because collinearity condition requires that one vector can be expressed as a scalar multiple of another, which holds for all values of k.

274 Medium

274 of 318

Understanding +4 | -1

\mathbf{a}

and

\mathbf{b}

are two non-collinear vectors and

x \mathbf{a}+y \mathbf{b}=0

View answer and explanation

Correct option: C

x=0, y=0

Option C (

x=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.

275 Easy

275 of 318

Application +4 | -1

If the position vectors of the points A, B and C be

\hat i + \hat j,\,\,\hat i - \hat j\,\,\,\,{\text{and}}\,\,\,\,a\hat i + b\hat j + c\hat k

respectively, then the points A, B and C are collinear, if

View answer and explanation

Correct option: D

c = 0, a = 1 and b is arbitrary scalars

Option D (c = 0, a = 1 and b is arbitrary scalars) is correct because for A, B, and C to be collinear, the vector AB must be a scalar multiple of the vector BC, leading to the condition c = 0 and a = 1.

276 Medium

276 of 318

Application +4 | -1

If two points are

P(7,-5,11)

and

Q(-2,8,13)

, then the projection of PQ on a straight line with direction cosines

\frac{1}{3}, \frac{2}{3}, \frac{2}{3}

View answer and explanation

Correct option: A

Option A (7) is correct because the projection is calculated using the direction cosines and the difference in coordinates of the two points.

277 Easy

277 of 318

Application +4 | -1

\mathbf{a}

and

\mathbf{b}

are non-collinear vectors, then the value of

\lambda

for which

\mathbf{u}=(\lambda+2) \mathbf{a}+\mathbf{b}

and

\mathbf{v}=(1+4 \lambda) \mathbf{a}-2 \mathbf{b}

are collinear is

View answer and explanation

Correct option: B

\frac{3}{2}

Option B (

\frac{3}{2}

) is correct because it is the value of

\lambda

for which

\mathbf{u}

and

\mathbf{v}

are collinear.

278 Easy

278 of 318

Application +4 | -1

The position vectors of the point

A\;\mathrm{and\;}B

with respect to

O

are

2\widehat \mathbf{i} + 2\widehat \mathbf{j} + \mathbf{\hat k}

and

2\widehat \mathbf{i} + 4\widehat \mathbf{j} + 4\mathbf{\hat k}

. The length of the internal bisector of

\angle \;BOA

\Delta AOB

View answer and explanation

Correct option: B

\frac{{\sqrt {136} }}{3}

Option B (

\frac{{\sqrt {136} }}{3}

) is correct because it represents the length of the internal bisector calculated using the section formula and vector properties.

279 Easy

279 of 318

Application +4 | -1

The resultant of

\left( {\mathbf{\vec p} - 2\mathbf{\vec q}} \right)

where.

\mathbf{\vec p} = 7\widehat \mathbf{i} - 2\widehat \mathbf{j} + 3\widehat {\mathbf{\dot k}}

and

\mathbf{\vec q} = 3\widehat \mathbf{i} + \widehat \mathbf{j} + 5\widehat {\mathbf{\dot k}}

View answer and explanation

Correct option: D

\sqrt {66}

Option D (

\sqrt {66}

) is correct because it represents the magnitude of the resultant vector after performing the scalar multiplication and vector subtraction.

280 Medium

280 of 318

Understanding +4 | -1

The unit vector bisecting OY and OZ is

View answer and explanation

Correct option: C

\frac{{\widehat j + \widehat k}}{{\sqrt 2 }}

Option C (

\frac{{\widehat j + \widehat k}}{{\sqrt 2 }}

) is correct because it is a unit vector with equal components along the OY and OZ axes.

281 Easy

281 of 318

Application +4 | -1

The perimeter of the triangle whose vertices have the position vectors

\mathbf{\hat i} + \mathbf{\hat j} + \mathbf{\hat k},\;5\mathbf{\hat i} + 3\mathbf{\hat j} - 3\mathbf{\hat k}

and

2\mathbf{\hat i} + 5\mathbf{\hat j} + 9\mathbf{\hat k}

is given by

View answer and explanation

Correct option: A

15 + \sqrt {157}

Option A (

15 + \sqrt {157}

) is correct because it is the sum of the lengths of all three sides of the triangle.

282 Medium

282 of 318

Application +4 | -1

A B C D

is a parallelogram and the position vectors of

A, B

and

C

are

\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}

and

7 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}

, then the position vector of

D

will be

View answer and explanation

Correct option: B

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

\frac{\mathbf{A} + \mathbf{C}}{2} = \frac{\mathbf{B} + \mathbf{D}}{2}

, which leads to

\mathbf{D} = \mathbf{A} + \mathbf{C} - \mathbf{B} = 7 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}

283 Easy

283 of 318

Application +4 | -1

The points with position vectors

10\hat i + 3\hat j,\;12\hat i - 5\hat j

and

a\hat i + 11\hat j

are collinear if the value of

a

View answer and explanation

Correct option: C

Option C (8) is correct because collinearity implies that one vector can be expressed as a scalar multiple of another.

284 Easy

284 of 318

Application +4 | -1

The points with position vectors

10\hat i + 3\hat j,\,\,12\hat i - 5\hat j\,\,\,{\text{and}}\,\,a\hat i + 11\hat j

are collinear, if a is equal to

View answer and explanation

Correct option: C

Option C (8) is correct because it represents the value of 'a' for which the given points are collinear, calculated using the condition for collinearity of points.

285 Easy

285 of 318

Application +4 | -1

If the position vector of

A

with respect to

O

3\widehat \mathbf{i} - 2\widehat \mathbf{j} + 4\mathbf{\hat k}

and

\overrightarrow {\mathbf{AB}} = 3\widehat \mathbf{i} - \widehat \mathbf{j} + \mathbf{\hat k}

Then the position vector of

B

with respect to

O

View answer and explanation

Correct option: B

6\widehat \mathbf{i} - 3\widehat \mathbf{j} + 5\mathbf{\hat k}

Option B (

6\widehat \mathbf{i} - 3\widehat \mathbf{j} + 5\mathbf{\hat k}

) is correct because it is the result of adding the position vector of A with respect to O and the vector AB.

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

286 Medium

286 of 318

Application +4 | -1

If the four points, whose position vectors are

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

5 \hat{i}-2 \alpha \hat{j}+4 \hat{k}

are coplanar, then

\alpha

is equal to

View answer and explanation

Correct option: C

\frac{73}{17}

Option C (

\frac{{73}}{{17}}

) is correct because it is the value of

\alpha

that makes the four points coplanar, calculated using the determinant condition for coplanarity.

287 Easy

287 of 318

Application +4 | -1

\mathbf{\vec a} = \widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k},\;\mathbf{\vec b} = \widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k}

and

\mathbf{\vec c} = x\;\widehat \mathbf{i} + \left( {x - 2} \right)\widehat \mathbf{j} - \mathbf{\hat k}

and if the vector

{\mathbf{\vec c}}

lies in the plane of vectors

{\mathbf{\vec a}}

and

\mathbf{\vec b},

then

x

equals

View answer and explanation

Correct option: C

- 2

Option C (-2) is correct because if

\mathbf{\vec c}

lies in the plane of

\mathbf{\vec a}

and

\mathbf{\vec b}

, then

\mathbf{\vec a}

\mathbf{\vec b}

, and

\mathbf{\vec c}

are coplanar, and their scalar triple product is zero, leading to x = -2.

288 Easy

288 of 318

Understanding +4 | -1

G

is the centroid of

\mathrm{\Delta }ABC

and

G'

is the centroid of

\mathrm{\Delta }A'B'C'

, then

A\vec A' + B\vec B' + C\vec C' =

View answer and explanation

Correct option: B

3G\vec G'

Option B (

3G\vec G'

) is correct because it is derived from expressing the sum of vectors

A\vec A' + B\vec B' + C\vec C'

in terms of the centroids G and G'.

289 Easy

289 of 318

Application +4 | -1

The equation of the plane containing the line

\mathbf{r}=\mathbf{i}+\mathbf{j}+\lambda(2 \mathbf{i}+\mathbf{j}+4 \mathbf{k})

View answer and explanation

Correct option: A

\mathbf{r}_{-}(\mathbf{i}+2 \mathbf{j}-\mathbf{k})=3

Option A is correct because the plane contains the given line, meaning any point on the line satisfies the equation of the plane.

290 Easy

290 of 318

Application +4 | -1

The magnitude of mutually perpendicular forces a, b and c are 2, 10 and 11respectively. Then the magnitude of its resultant is

View answer and explanation

Correct option: B

Option B (15) is correct because the magnitude of the resultant of three mutually perpendicular forces is the square root of the sum of the squares of their magnitudes.

291 Easy

291 of 318

Application +4 | -1

The direction cosines of vector

a = 3\hat i + 4\hat j + 5\hat k

in the direction of positive axis of X, is

View answer and explanation

Correct option: C

\frac{3}{{\sqrt {50} }}

Option C (

\frac{3}{{\sqrt {50} }}

) is correct because it represents the direction cosine of the vector along the positive X-axis, calculated as the x-component divided by the magnitude of the vector.

292 Easy

292 of 318

Application +4 | -1

O P=8

and

\overrightarrow{O P}

makes angles

45^{\circ}

and

60^{\circ}

with

O X

-axis and

O Y

-axis respectively, then

\overrightarrow{O P}=

View answer and explanation

Correct option: B

4(\sqrt{2} \mathbf{i}+\mathbf{j} \pm \mathbf{k})

Option B (

4(\sqrt{2} \mathbf{i}+\mathbf{j} \pm \mathbf{k})

) is correct because it is the only vector with a magnitude of 8 that satisfies the given angle conditions.

293 Easy

293 of 318

Analyzing +4 | -1

If the vectors

\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}-2 \hat{k}

and

\vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}

are coplanar and the projection of

\vec{a}

on the vector

\vec{b}

\sqrt{54}

units, then the sum of all possible values of

\lambda+\mu

is equal to

View answer and explanation

Correct option: A

Option A (24) is correct because it is the sum of all possible values of

\lambda + \mu

calculated from the coplanarity condition and the projection of vectors.

294 Easy

294 of 318

Application +4 | -1

\mathbf{\vec a} = \mathbf{\hat i} + \mathbf{\hat j} + \mathbf{\hat k},\;\mathbf{\vec b} = 2\mathbf{\hat i} - 4\mathbf{\hat k},\;\mathbf{\vec c} = \mathbf{\hat i} + \mathrm{\lambda }\mathbf{\hat j} + 3\mathbf{\hat k}

are coplanar, then the value of

\lambda

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because the calculated value of lambda is 5/3, which is not among the given options.

295 Easy

295 of 318

Application +4 | -1

If the vectors

\mathbf{\vec a} = 2\widehat \mathbf{i} + 3\widehat \mathbf{j} + 6\mathbf{\hat k}

and

{\mathbf{\vec b}}

are collinear and

\left| {\mathbf{\vec b}} \right| = 21,\;\mathrm{then\;}\mathbf{\vec b}

is equal to

View answer and explanation

Correct option: B

\pm 3\left( {2\widehat \mathbf{i} + 3\widehat \mathbf{j} + 6\mathbf{\hat k}} \right)

Option B (

\pm 3\left( {2\widehat \mathbf{i} + 3\widehat \mathbf{j} + 6\mathbf{\hat k}} \right)

) is correct because collinear vectors are scalar multiples of each other, and the magnitude condition determines the scalar.

296 Easy

296 of 318

Application +4 | -1

Let

\mathbf{r}=\mathbf{a}+\lambda \mathbf{I}

and

\mathbf{r}=\mathbf{b}+\mu \mathrm{m}

be two lines in space, where

\mathbf{a}=5 \mathbf{i}+\mathbf{j}+2 \mathbf{k}, \mathbf{b}=-\mathbf{i}+7 \mathbf{j}+8 \mathbf{k}, I=-4 \mathbf{i}+\mathbf{j}-\mathbf{k}

, and

\mathbf{m}=2 \mathbf{i}-5 \mathbf{j}-7 \mathbf{k}

, then the position vector of a point which lies on both of these lines, is

View answer and explanation

Correct option: A

\mathbf{i}+2 \mathbf{j}+\mathbf{k}

Option D (non-existent as the lines are skew) is correct because the lines do not intersect, meaning there is no common point.

297 Easy

297 of 318

Application +4 | -1

\widehat \mathbf{i} + 2\widehat \mathbf{j} + 3\mathbf{\hat k}

and

2\widehat \mathbf{i} - \widehat \mathbf{j} + 4\mathbf{\hat k}

are the position vectors of the points A and B , then the position vector of the points of trisection of AB are

View answer and explanation

Correct option: A

\frac{4}{3}\widehat \mathbf{i} + \widehat \mathbf{j} + \frac{{10}}{3}\mathbf{\hat k},\frac{5}{3}\widehat \mathbf{i} + \frac{{11}}{3}\mathbf{\hat k}

Option A (

\frac{4}{3}\widehat{\mathbf{i}} + \widehat{\mathbf{j}} + \frac{{10}}{3}\mathbf{\hat k},\frac{5}{3}\widehat{\mathbf{i}} + \frac{{11}}{3}\mathbf{\hat k}

) is correct because it represents the position vectors of the points that trisect the line segment AB, calculated using the section formula.

298 Easy

298 of 318

Application +4 | -1

\vec{a}=(1,-1,2), \vec{b}=(-2,3,5)

\vec{c}=(2,-2,4)

and

\hat{i}

is the unit vector in the

x

-direction, then

(\vec{a}-2 \vec{b}+3 \vec{c}) \hat{i}=

View answer and explanation

Correct option: A

Option A (11) is correct because after substituting the vectors and performing the dot product with

\hat{i}

, the result is 11.

299 Easy

299 of 318

Application +4 | -1

ABCDE is a pentagon. Forces AB, AE, DC and ED act at a point. Which force should be added to this system to make the resultant 2AC?

View answer and explanation

Correct option: C

Option C (BC) is correct because adding BC to the given forces results in a vector sum of 2AC, as derived from vector addition properties.

300 Easy

300 of 318

Application +4 | -1

The value of

\lambda

for which the four points

2\hat i + 3\hat j - \hat k,\,\,\hat i + 2\hat j + 3\hat k,\,\,3\hat i + 4\hat j - 2\hat k\,\,\,{\text{and}}\,\,\,\hat i - \lambda \hat j + 6\hat k\,

are coplanar

View answer and explanation

Correct option: C

-2

Option C (-2) is correct because it satisfies the condition for coplanar points, which can be determined using the determinant of the matrix formed by the position vectors.

Revision notes and key takeaways

Review ADDITION OF VECTORS in your textbook or class notes, then use this set to check your understanding.

Showing 271-300 of 318 questions

Your Progress

Addition of vectors practice · Mixed difficulty

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice