If ${\bf{\vec a}} = \left( {2,1, - 1} \right),\;{\bf{\vec b}} = \left( {1, - 1,0} \right),\;{\bf{\vec c}} = \left( {5 - 1,1} \right),\;$ then unit vector parallel to ${\bf{\vec a}} + {\bf{\vec b}} - {\bf{\vec c}}$ but in opposite direction is

$\frac{1}{3}\left( {2\widehat {\bf{i}} - \widehat {\bf{j}} + 2{\bf{\hat k}}} \right)$

$\frac{1}{2}\left( {2\widehat {\bf{i}} - \widehat {\bf{j}} + 2{\bf{\hat k}}} \right)$

$\frac{1}{3}\left( {2\widehat {\bf{i}} - \widehat {\bf{j}} - 2{\bf{\hat k}}} \right)$

If ${\bf{\vec a}},\,\,{\bf{\vec b}},\,\,{\bf{\vec c}}\;$ are three non-coplanar vectors then the vector equation ${\bf{\vec r}} = \left( {1 - p - q} \right)\,\,{\bf{\vec a}} + p\,{\bf{\vec b}} + q\,{\bf{\vec c}}$ represent a

Plane passing through the origin

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JEE • MATHEMATICS • ADDITION OF VECTORS

Addition of vectors practice · Mixed difficulty

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181 Easy

181 of 318

Application +4 | -1

If a, b, c are three non-coplanar vectors, then

3a - 7b - 4c,\,\,\,3a - 2b + c

and

a + b + \lambda c

will be coplanar, if

\lambda

View answer and explanation

Correct option: D

Option D (2) is correct because for the vectors to be coplanar, their scalar triple product must be zero, which leads to λ = 2.

182 Easy

182 of 318

Analyzing +4 | -1

Let

a, b, c

be distinct non-negative numbers. If the vectors

a \hat{\mathbf{i}}+a \hat{\mathbf{j}}+c \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}

and

c \hat{\mathbf{i}}+c \hat{\mathbf{j}}+b \hat{\mathbf{k}}

lie in a plane, then

c

View answer and explanation

Correct option: D

GM of

a

and

b

Option D (GM of

a

and

b

) is correct because the condition for the vectors to lie in a plane leads to the relationship

c^2 = ab

, which defines

c

as the geometric mean of

a

and

b

183 Easy

183 of 318

Application +4 | -1

Let

a = \hat i

be a vector which makes an angle of

120^\circ

with a unit vector b. Then, the unit vector

(a + b)

View answer and explanation

Correct option: C

\frac{1}{2}\hat i + \frac{{\sqrt 3 }}{2}\hat j

Option C (

\frac{1}{2}\hat i + \frac{{\sqrt 3 }}{2}\hat j

) is correct because it represents the unit vector (a + b) calculated using the given angle and the properties of unit vectors.

184 Easy

184 of 318

Analyzing +4 | -1

{a_1}

and

{a_2}

are two values of a for which the unit vector

a\widehat i + b\widehat j + \frac{1}{2}\widehat k

is linearly dependent with

\widehat i + 2\widehat j

and

\widehat j - 2\widehat k,

then

\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}}

is equal to

View answer and explanation

Correct option: C

\frac{{ - 16}}{{11}}

Option D (

\frac{{ - 16}}{{11}}

) is correct because it is the value of

\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}}

calculated from the quadratic equation derived from the given conditions.

185 Easy

185 of 318

Application +4 | -1

If the position vectors of P and Q are

\hat i + 2\hat j - 7\hat k\,\,\,\,{\text{and}}\,\,\,5\hat i - 3\hat j + 4\hat k\,

respectively, the cosine angle between PQ and Z – axis is

View answer and explanation

Correct option: B

\frac{{11}}{{\sqrt {162} }}

Option B (

\frac{{11}}{{\sqrt {162} }}

) is correct because it represents the cosine of the angle between PQ and the Z-axis.

186 Easy

186 of 318

Application +4 | -1

If the

p \hat{i}+\hat{j}+\hat{k}, \hat{i}+q \hat{j}+\hat{k}

and

\hat{i}+\hat{j}+r \hat{k}(p \neq q \neq r \neq 1)

vector are coplanar, then the value of

p q r-(p+q+r)

View answer and explanation

Correct option: D

-2

Option D (-2) is correct because if the given vectors are coplanar, the determinant formed by their components is zero. Solving this determinant equation leads to the result

pqr - (p+q+r) = -2

187 Medium

187 of 318

Analyzing +4 | -1

|a + b| < |a - b|,

then the angle between a and b can lie in the interval.

View answer and explanation

Correct option: C

(\pi /2,\,\,3\pi /2)

Option C (

\pi /2,\,3\pi /2

\)) is correct because

|a + b| < |a - b|

implies that the angle between a and b is obtuse, lying between 90 and 270 degrees.

188 Medium

188 of 318

Application +4 | -1

The image of the point having the position vector

\hat{i}+3 \hat{j}+4 \hat{k}

in the plane

\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0

View answer and explanation

Correct option: C

-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}

Option C (

-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}

) is correct. By finding the equation of the line passing through the given point and normal to the plane, we can find the image point.

189 Easy

189 of 318

Understanding +4 | -1

The vector equation of the symmetrical form of equation of straight line

\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}

View answer and explanation

Correct option: D

\overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})+\mu(3 \hat{\mathrm{i}}+7 \mathrm{j}+2 \hat{\mathrm{k}})

Option D (

\overrightarrow{\mathrm{{r}}}=(5 \hat{\mathrm{{i}}}-4 \hat{\mathrm{{j}}}+6 \hat{\mathrm{{k}}})+\mu(3 \hat{\mathrm{{i}}}+7 \hat{\mathrm{{j}}}+2 \hat{\mathrm{{k}}})

) is correct because it correctly represents the vector equation of a line passing through the point (5, -4, 6) with direction ratios (3, 7, 2).

190 Easy

190 of 318

Application +4 | -1

The vectors

2\,\hat i + 3\hat j,\;\,5\,\hat i + 6\hat j

and

8\,\hat i + \lambda \hat j

have their initial points at

\left( {\mathrm{1, 1}} \right)

. The value of

\lambda

so that the vectors terminate on one straight line, is

View answer and explanation

Correct option: D

Option D (9) is correct because the value of

\lambda

is found by ensuring that the terminal points of the given vectors are collinear, which leads to a linear equation that can be solved for

\lambda

191 Easy

191 of 318

Application +4 | -1

The position vector of the points which divides internally in the ratio 2 : 3 the join of the points 2a – 3b and 3a – 2b is

View answer and explanation

Correct option: B

\frac{{12}}{5}a - \frac{{13}}{5}b

Option B (

\frac{{12}}{5}a - \frac{{13}}{5}b

) is correct because it represents the position vector of the point that divides the line segment in the given ratio, calculated using the section formula.

192 Easy

192 of 318

Application +4 | -1

The ratio in which

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

divides the join of

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

and

7\widehat \mathbf{i} - \mathbf{\hat k}

View answer and explanation

Correct option: A

2:1

The correct answer is A. The ratio is found by using the section formula and equating the coefficients of the vectors.

193 Medium

193 of 318

Application +4 | -1

The value of

k

for which the points

A(1,0,3)

B(-1,3,4), C(1,2,1)

and

D(k, 2,5)

are coplanar, are

View answer and explanation

Correct option: D

-1

Option D (-1) is correct because the value of k is found by setting the scalar triple product of the vectors formed by the points to zero, indicating coplanarity.

194 Easy

194 of 318

Application +4 | -1

If the position vectors of the points

A

and

B

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

and

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

, then the line

A B

is parallel to

View answer and explanation

Correct option: C

2 x

-plane

Option C (

2 x

-plane) is correct because the vector

\overrightarrow{A B}

has no

\mathbf{j}

component, indicating it lies in a plane parallel to the

xz

-plane.

195 Easy

195 of 318

Application +4 | -1

Equation of the line which passes through the point with position vector

(2,1,0)

and perpendicular to the plane containing the vectors

\mathbf{i}+\mathbf{j}

and

\mathbf{j}+\mathbf{k}

is
Where,

t

is a parameter.

View answer and explanation

Correct option: A

\mathrm{r}=(2,1,0)+t(1,-1,1)

Option A (

\mathrm{r}=(2,1,0)+t(1,-1,1)

) is correct because it represents a line passing through (2,1,0) and perpendicular to the plane defined by the cross product of the given vectors.

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

196 Medium

196 of 318

Application +4 | -1

\mathbf{a}, \mathbf{b}, \mathbf{c}

are the three non-coplanar vectors and

\mathbf{p}, \mathbf{q}, \mathbf{r}

are defined by the relations

p=\frac{b \times c}{[a b c]}, q=\frac{c \times a}{[a b c]}, r=\frac{a \times b}{[a b c]}

then

(a+b) \cdot p+(b+c)

\mathbf{q}+(\mathbf{c}+\mathbf{a}) \cdot \mathbf{r}=

View answer and explanation

Correct option: D

Option D (3) is correct because by using the properties of scalar triple product and reciprocal vectors, the expression simplifies to 3.

197 Easy

197 of 318

Analyzing +4 | -1

If the vectors

\vec a = \hat i + a\hat j + {a^2}\hat k,\vec b = \hat i + b\hat j + {b^2}\hat k,\vec c = \hat i + c\hat j + {c^2}\hat k

are three non-coplanar vectors and

\left| {\begin{array}{cccccccccccccccccccc} a&{{a^2}}&{1 + {a^3}}\\ b&{{b^2}}&{1 + {b^3}}\\ c&{{c^2}}&{1 + {c^3}} \end{array}} \right| = 0,

then the value of

abc

View answer and explanation

Correct option: D

- 1

The correct answer is D (-1) because the given determinant can be factored into the product of another determinant and (1 + abc), and since the vectors are non-coplanar, the determinant must be -1.

198 Hard

198 of 318

Analyzing +4 | -1

If the points

P

and

Q

are respectively the circumcenter and the orthocentre of a

\triangle A B C

, then

\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}

is equal to

View answer and explanation

Correct option: C

\overrightarrow{P Q}

Option C (

\overrightarrow{{PQ}}

) is correct because the vector sum

\overrightarrow{{PA}} + \overrightarrow{{PB}} + \overrightarrow{{PC}}

is equal to

\overrightarrow{{PQ}}

based on the properties of the circumcenter, orthocenter, and centroid of a triangle.

199 Easy

199 of 318

Analyzing +4 | -1

(x,y,z) \ne (0,0,0)

and

(\widehat i + \widehat j + 3\widehat k)x + (3\widehat i - 3\widehat j + \widehat k)y + ( - 4\widehat i + 5\widehat j)z = \lambda (x\widehat i + y\widehat j + z\widehat k),

then the value of

\lambda

will be

View answer and explanation

Correct option: D

0, -1

Option D (0, -1) is correct because these are the values of

\lambda

for which the given equation holds true, derived by comparing coefficients and solving the determinant.

200 Easy

200 of 318

Application +4 | -1

Given three vectors

a = 6\hat i - 3\hat j,\,\,\,\,b = 2\hat i - 6\hat j{\text{ }}\,\,{\text{and }}\,\,c = - 2\hat i + 21\hat j

such that

\alpha = a + b + c.

Then, the resolution of the vector

\alpha

into components with respect to a and b is given by

View answer and explanation

Correct option: C

2a - 3b

Option C (2a - 3b) is correct because it represents the resolution of vector α into components with respect to vectors a and b, calculated by solving a system of equations.

201 Easy

201 of 318

Application +4 | -1

The value of

k

for which the vectors

\mathbf{a}=\mathbf{i}-\mathbf{j}

and

\mathbf{b}=-2 \mathbf{i}+k \mathbf{j}

are collinear is

View answer and explanation

Correct option: A

Option A (2) is correct because for the vectors to be collinear, one must be a scalar multiple of the other, leading to k=2.

202 Easy

202 of 318

Application +4 | -1

The vectors

\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}, \quad \lambda \mathbf{i}+4 \mathbf{j}+7 \mathbf{k}, \quad-3 \mathbf{i}-2 \mathbf{j}-5 \mathbf{k}

are collinear, if

\lambda

equals

View answer and explanation

Correct option: A

Option A (3) is correct because the determinant of the matrix formed by the collinear vectors must be zero, leading to

\lambda = 3

203 Easy

203 of 318

Application +4 | -1

{a}=2 {i}-{j}+{k}, {b}={i}+2 {j}-3 {k}

and

{c}=3 {i}+\mu {j}+5 {k}

are coplanar, then

\mu

is a roots of the equation

View answer and explanation

Correct option: A

x^2+3 x=4

Option A (

x^2+3 x=4

) is correct because

\mu = -4

is a root of this equation, derived from the coplanarity condition.

204 Medium

204 of 318

Application +4 | -1

\mathbf{\vec a} = \left( {2,1, - 1} \right),\;\mathbf{\vec b} = \left( {1, - 1,0} \right),\;\mathbf{\vec c} = \left( {5 - 1,1} \right),\;

then unit vector parallel to

\mathbf{\vec a} + \mathbf{\vec b} - \mathbf{\vec c}

but in opposite direction is

View answer and explanation

Correct option: A

\frac{1}{3}\left( {2\widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k}} \right)

Option A (

\frac{1}{3}\left( {2\widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k}} \right)

) is correct because it accurately calculates the unit vector parallel to

\mathbf{\vec a} + \mathbf{\vec b} - \mathbf{\vec c}

but in the opposite direction.

205 Easy

205 of 318

Application +4 | -1

If the position vectors of the points A and B are

\hat i + 3\hat j - \hat k\,\,\,{\text{and}}\,\,\,3\hat i - \hat j - 3\hat k

, , then what will be the position vector of the mid – point of AB

View answer and explanation

Correct option: B

2\hat i + \hat j - 2\hat k

Option B (

2\hat i + \hat j - 2\hat k

) is correct because it represents the position vector of the midpoint of AB, calculated by averaging the position vectors of A and B.

206 Medium

206 of 318

Application +4 | -1

Let

\overrightarrow{\mathrm{OA}}=\vec{a}, \overrightarrow{\mathrm{OB}}=2 \vec{a}+10 \vec{b}, \overrightarrow{O C}=\vec{b}

where

\mathrm{O}, \mathrm{A}, \mathrm{C}

are non collinear points. Let

\lambda

denote the area of the quadrilateral

\mathrm{OABC}

. Let

\mathrm{m}

denotes the area of parallelogram with

\overrightarrow{\mathrm{OA}}

and

\overrightarrow{\mathrm{OC}}

as adjacent sides. If

\lambda=2 \lambda \mathrm{m}

then find the value of

\lambda

View answer and explanation

Correct option: A

Option A (3) is correct because by calculating the areas of the quadrilateral and parallelogram and using the given relation, the value of lambda is found to be 3.

207 Easy

207 of 318

Understanding +4 | -1

\mathbf{\vec a},\,\,\mathbf{\vec b},\,\,\mathbf{\vec c}\;

are three non-coplanar vectors then the vector equation

\mathbf{\vec r} = \left( {1 - p - q} \right)\,\,\mathbf{\vec a} + p\,\mathbf{\vec b} + q\,\mathbf{\vec c}

represent a

View answer and explanation

Correct option: B

Plane

Option B (Plane) is correct because the given vector equation represents a plane since the coefficients of the vectors sum to 1.

208 Medium

208 of 318

Application +4 | -1

A unit vector in the plane of the vectors

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

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

and orthogonal to

5 \mathbf{i}+2 \mathbf{j}+6 \mathbf{k}

View answer and explanation

Correct option: B

\frac{3 \mathrm{j}-\mathbf{k}}{\sqrt{10}}

Option B (

\frac{{3 \mathrm{j}-\mathbf{k}}}{{\sqrt{{10}}}}

) is correct because it satisfies the conditions of being in the plane of the given vectors and orthogonal to the specified vector.

209 Medium

209 of 318

Application +4 | -1

If the points whose position, vectors are

3 \mathbf{i}-2 \mathbf{j}-\mathbf{k}

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

and

4 \mathbf{i}+5 \mathbf{j}+\lambda \mathbf{k}

lie on a plane, then

\lambda=

View answer and explanation

Correct option: A

-\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.

210 Easy

210 of 318

Application +4 | -1

the area of the parallelogram whose adjacent sides are

\mathbf{i}-\mathbf{k}

and

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

View answer and explanation

Correct option: C

\sqrt{17}

Option C (

\sqrt{{17}}

) is correct because it's the magnitude of the cross product of the adjacent sides of the parallelogram.

Revision notes and key takeaways

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

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