Let $\mathbf{b}=-\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ and $\mathbf{c}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}$. If $\mathbf{a}$ be $a$ unit vector and the scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$ has the greatest value, then a is equal to

$\frac{1}{\sqrt{3}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$

$\frac{1}{\sqrt{5}}(\sqrt{2} \hat{i}-\hat{j}-\sqrt{2} \hat{k})$

$\frac{1}{\sqrt{59}}(3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}})$

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JEE • MATHEMATICS • SCALAR TRIPLE PRODUCT

Scalar triple product practice · Medium

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Showing 1-30 of 38 questions

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1 Medium

1 of 38

Application +4 | -1

The length of the perpendicular from the origin to the plane passing through the point a and containing the line

\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}

View answer and explanation

Correct option: C

\frac{[a \mathbf{b c}]}{|\mathbf{b} \times \mathbf{c}+\mathbf{c \times a}|}

The correct answer is

\frac{{[a \mathbf{b c}]}}{{|\mathbf{b} \times \mathbf{c}+\mathbf{c \times a}|}}

because it represents the formula for the perpendicular distance from the origin to the plane.

2 Medium

2 of 38

Application +4 | -1

Let

\mathbf{b}=-\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}

and

\mathbf{c}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}

. If

\mathbf{a}

a

unit vector and the scalar triple product

[\mathbf{a} \mathbf{b} \mathbf{c}]

has the greatest value, then a is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the scalar triple product is maximized when a is in the direction of b x c, and normalizing b x c gives the unit vector a.

3 Medium

3 of 38

Application +4 | -1

The vector equation of a plane, which is at a distance of 8 unit from the origin and which is normal to the vector

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

, is

View answer and explanation

Correct option: B

\mathbf{r} \cdot(2 \mathbf{i}+\mathbf{j}+2 \mathbf{k})=24

Option B,

\mathbf{r} \cdot(2 \mathbf{i}+\mathbf{j}+2 \mathbf{k})=24

, is correct because it represents the vector equation of a plane at a distance of 8 units from the origin and normal to the given vector.

4 Medium

4 of 38

Analyzing +4 | -1

Let

\mathbf{r}=(\mathbf{a} \times \mathbf{b}) \sin x+(\mathbf{b} \times \mathbf{c}) \cos y+(\mathbf{c} \times \mathbf{a})

, where

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

are non-zero non-coplanar vectors. If

\mathbf{r}

is orthogonal to

3 \mathbf{a}+5 \mathbf{b}+2 \mathbf{c}

, then the value of

\sec ^2 y+\operatorname{cosec}^2 x+\sec y \operatorname{cosec} x

View answer and explanation

Correct option: A

Option A (3) is correct because substituting

\sin x = -1

and

\cos y = -1

into the expression

\sec^2 y + \csc^2 x + \sec y \csc x

yields 3.

5 Medium

5 of 38

Application +4 | -1

The shortest distance between the lines

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

and

\mathbf{r}=\mathbf{i}-\mathbf{j}+2 \mathbf{k}+\mathbf{j} s

(

t

and

s

being parameters) is

View answer and explanation

Correct option: C

Option C (4) is correct because the shortest distance between the two lines is calculated using the formula and found to be 4.

6 Medium

6 of 38

Application +4 | -1

The shortest distance between the lines

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

and

\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}

View answer and explanation

Correct option: D

3 \sqrt{6}

Option D (

3 \sqrt{6}

) is correct because it is the shortest distance calculated using the formula for the distance between two skew lines.

7 Medium

7 of 38

Analyzing +4 | -1

Let a vector

a

be coplanar with vectors

b = 2\hat i + \hat j + \hat k

and

c = \hat i - \hat j + \hat k

. If

a

is perpendicular to

d = 3\hat i + 2\hat j + 6\hat k

and

\left| a \right| = \sqrt {10}

. Then possible value of

\left[ {\begin{array}{cccccccccccccccccccc} a&b&c \end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc} a&b&d \end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc} a&c&d \end{array}} \right]

is equal to

View answer and explanation

Correct option: A

-42

Option A (-42) is correct because given the conditions, the scalar triple product can be computed using the properties of coplanar vectors and perpendicularity, resulting in -42.

8 Medium

8 of 38

Analyzing +4 | -1

Let

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

be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and

(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=168

, then

|\vec{a}|+|\vec{b}|+|\vec{c}|

is equal to :

View answer and explanation

Correct option: C

Option D (

\frac{95}{8}

) is the correct answer because it is the product of the roots of the equation obtained from the coplanarity condition.

9 Medium

9 of 38

Analyzing +4 | -1

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

are non-coplanar vectors and

\left( {\mathbf{\vec a} - \mathrm{\lambda }\;\mathbf{\vec b}} \right)\cdot\left( {\mathbf{\vec b} - 2\mathbf{\vec c}} \right) \times \left( {\mathbf{\vec c} + 2\mathbf{\vec a}} \right) = 0,

then λ is equal to

View answer and explanation

Correct option: D

- 1/4

The correct answer is D (

- 1/4

) because the scalar triple product is zero, implying coplanarity, which leads to the solution λ = -1/4.

10 Medium

10 of 38

Analyzing +4 | -1

\alpha

and

\beta

are two mutually perpendicular unit vectors

\{r \alpha+r+s(\alpha \times \beta\},[\alpha+(\alpha \times \beta)]

and

\{s \alpha+s \beta+t(\alpha \times \beta)\}

are coplanar, then

s

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the condition for coplanarity leads to

s^2 = rt

, implying

s

is the geometric mean of

r

and

t

, but the options do not explicitly state this.

11 Medium

11 of 38

Analyzing +4 | -1

Let area of faces,

\triangle O A B=\lambda_1, \Delta O A C=\lambda_2, \Delta O B C=\lambda_3, \Delta A B C=\lambda_4

and

h_1, h_2, h_3, h_4

be perpendicular height from 0 to face

\triangle A B C, A

to the face

\triangle O B C, B

to the face

\triangle O A C, C

to the face

\triangle O A B

, then the face

\frac{1}{3} \lambda_1 h_4 \cdot \frac{1}{3} \lambda_2 h_3+\frac{1}{3} \lambda_3 h_2+\frac{1}{3} \lambda_4 h_1

View answer and explanation

Correct option: A

\frac{2}{3}|[AB\,\,AC\,\,OA]|

Option A is correct because the sum of the volumes of the tetrahedra formed by dropping perpendiculars from the origin to each face equals twice the volume of the tetrahedron.

12 Medium

12 of 38

Application +4 | -1

|{a}|^2=8

and

{a} \times({i}+{j}+2 {k})={0}

then the value of

{a} \cdot(-{i}+{j}+4 {k})

View answer and explanation

Correct option: B

\frac{16}{\sqrt{3}}

Option B (

\frac{{16}}{{\sqrt{3}}}

) is correct because it is derived from the given condition a x (i + j + 2k) = 0 and |a|^2 = 8.

13 Medium

13 of 38

Analyzing +4 | -1

|\bar{a}|=1,|\bar{b}|=2,|\bar{c}|=3

and

\bar{a}, \bar{b}

are perpendicular to

\bar{c}

and

\bar{a} \cdot \bar{b}=1,(\bar{a}, \bar{b}, \bar{c}

form a vector traid in R.H.S) then

\left[\frac{\bar{a} \times \bar{b}}{5} \quad \bar{b} \times \bar{c} \frac{\bar{a}+\bar{b}+\bar{c}}{\sqrt{3}}\right]=

View answer and explanation

Correct option: C

Option C (3) is correct because the scalar triple product simplifies to 3 after applying vector identities and given conditions.

14 Medium

14 of 38

Understanding +4 | -1

[\mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]

is equal to

View answer and explanation

Correct option: C

[\mathbf{a} \mathbf{b} \mathbf{c}]^2

Option B is correct because it correctly expresses a complex number on the unit circle in terms of the tangent of half its argument.

15 Medium

15 of 38

Application +4 | -1

The shortest distance between the lines

\mathbf{r}_1=4 \mathbf{i}-3 \mathbf{j}-\mathbf{k}+\lambda(\mathbf{i}-4 \mathbf{j}+7 \mathbf{k})

and

\mathbf{r}_2=\mathbf{i}-\mathbf{j}-10 \mathbf{k}+\lambda(2 \mathbf{i}-3 \mathbf{j}+8 \mathbf{k})

View answer and explanation

Correct option: D

Option D (0) is correct because the shortest distance between the lines is calculated to be 0, indicating that the lines intersect.

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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

16 Medium

16 of 38

Analyzing +4 | -1

The equation of the plane containing the straight line

\frac{x}{2} = \frac{y}{3} = \frac{z}{4}

and perpendicular to the plane containing the straight lines

\frac{x}{3} = \frac{y}{4} = \frac{z}{2}

and

\frac{x}{4} = \frac{y}{2} = \frac{z}{3}

View answer and explanation

Correct option: D

x - 2y + z = 0

Option D (

x - 2y + z = 0

) is correct because it represents the equation of the plane containing the given line and perpendicular to the plane containing the other two lines, obtained by finding the normal vector and using the point-normal form.

17 Medium

17 of 38

Analyzing +4 | -1

Let

\hat{\mathbf{a}}

and

\hat{\mathbf{b}}

be two unit vectors such that

\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}=\frac{1}{3}

and

\hat{\mathbf{a}} \times \hat{\mathbf{b}}=\hat{\mathbf{c}}

. Also

\bar{F}=\alpha \hat{\mathbf{a}}+\beta \hat{\mathbf{b}}+\gamma \hat{\mathbf{c}}

, where

\alpha, \beta, \gamma

are scalars. If

\alpha=k_1(\hat{\mathbf{F}} \cdot \hat{\mathbf{a}})-k_2(\hat{\mathbf{F}} \cdot \hat{\mathbf{b}})

, then the value of

2\left(k_1+k_2\right)

View answer and explanation

Correct option: C

The correct answer is C (3) because it is obtained by calculating the scalar triple product and using vector identities to simplify the expression for alpha.

18 Medium

18 of 38

Application +4 | -1

The shortest distance between the lines

\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}

and

\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}

View answer and explanation

Correct option: D

Option D (9) is correct because it involves finding the shortest distance between two lines using the formula involving the cross product of direction vectors and the difference of position vectors.

19 Medium

19 of 38

Analyzing +4 | -1

If the vectors

\vec{a}

and

\vec{b}

are perpendicular to each other, then a vector

\vec{v}

in terms of

\vec{a}

and

\vec{b}

satisfying the equations

\vec{v} \cdot \vec{a}=0, \vec{v} \cdot \vec{b}=1

and

\left[\begin{array}{lll}\vec{v} & \vec{a} & \vec{b}\end{array}\right]=1,

View answer and explanation

Correct option: A

\frac{\vec{b}}{|\vec{b}|^2}+\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|^2}

Option A (

\frac{\vec{{b}}}{{|\vec{{b}}|^2}}+\frac{{\vec{{a}} \times \vec{{b}}}}{{|\vec{{a}} \times \vec{{b}}|^2}}

) is correct because it satisfies the given conditions for the vector

\vec{{v}}

when

\vec{{a}}

and

\vec{{b}}

are perpendicular.

20 Medium

20 of 38

Analyzing +4 | -1

\alpha {\text{ }}\,\,{\text{and}}\,\,{\text{ }}\beta

are two mutually perpendicular unit vectors

\left\{ {r\alpha + r\beta + s(\alpha \times \beta } \right\},\,\,\,[\alpha + (\alpha \times \beta )]

and

\left\{ {s\alpha + s\beta + t(\alpha \times \beta )} \right\}

are coplanar, then s is equal to

View answer and explanation

Correct option: B

GM of r and t

Option B (GM of r and t) is correct because the coplanarity condition leads to the equation s² = rt, which defines the geometric mean.

21 Medium

21 of 38

Application +4 | -1

Given vectors

\vec x = 3\hat i - 6\hat j - \hat k,\;\vec y = \hat i + 4\hat j - 3\hat k

and

\vec z = 3\hat i + 4\hat j + 12\hat k

, then the projection of

\vec x \times \vec y

on vector

{\vec z}

View answer and explanation

Correct option: B

- 14

Option B (

-14

) is correct because the projection is calculated using the scalar triple product divided by the magnitude of vector z.

22 Medium

22 of 38

Application +4 | -1

The shortest distance between the lines

\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}

and

\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}

View answer and explanation

Correct option: A

4 \sqrt{3}

Option A (

4 \sqrt{3}

) is correct because it is the shortest distance calculated using the formula for skew lines.

23 Medium

23 of 38

Application +4 | -1

Equation of the plane through three points

A, B

and

C

with position vectors

-6 \mathbf{i}+3 \mathbf{j}+2 \mathbf{k}, 3 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}

and

5 \mathbf{i}+7 \mathbf{j}+3 \mathbf{k}

is equal to

View answer and explanation

Correct option: A

\mathbf{r} \cdot(\mathbf{i}-\mathbf{j}+7 \mathbf{k})+23=0

Option A (

\mathbf{r} \cdot(\mathbf{i}-\mathbf{j}+7 \mathbf{k})+23=0

) is correct because it represents the equation of the plane passing through the three given points.

24 Medium

24 of 38

Analyzing +4 | -1

Let

\vec{a}=4 \hat{i}+3 \hat{j}

and

\vec{b}=3 \hat{i}-4 \hat{j}+5 \hat{k}

. If

\vec{c}

is a vector such that

\vec{c} \cdot(\vec{a} \times \vec{b})+25=0, \vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=4

and projection of

\vec{c}

\vec{a}

is 1 , then the projection of

\vec{c}

\vec{b}

equals

View answer and explanation

Correct option: B

\frac{5}{\sqrt{2}}

Option B (

\frac{5}{\sqrt{2}}

) is correct because it is the value obtained by correctly solving the system of equations and then calculating the projection of c on b.

25 Medium

25 of 38

Application +4 | -1

Let

\bar{u}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \bar{v}=3 \hat{i}+4 \hat{j}+5 \hat{k}

and

\bar{w}

be a unit vector, then the maximum value of the scalar triple product

[\bar{u} \bar{v} \bar{w}]

, is

View answer and explanation

Correct option: C

\sqrt{6}

Option C (

\sqrt{6}

) is correct because the maximum value of the scalar triple product occurs when

\bar{w}

is in the direction of

\bar{u} \times \bar{v}

, and its magnitude is

|\bar{u} \times \bar{v}|

26 Medium

26 of 38

Application +4 | -1

The equation

|\mathbf{r}|^2-\mathbf{r} .(2 \mathbf{i}+4 \mathbf{j}-2 \mathbf{k})-10=0

represents a

View answer and explanation

Correct option: C

Sphere of radius 4

Option C (Sphere of radius 4) is correct because the given equation represents a sphere, and its radius is calculated to be 4.

27 Medium

27 of 38

Application +4 | -1

The length of the perpendicular from the origin to the plane passing through three non-collinear points a,b,c is

View answer and explanation

Correct option: A

\frac{[a \mathbf{b} \mathbf{c}]}{|\mathbf{a} \times \mathbf{b}+\mathbf{c} \times \mathbf{a}+\mathbf{b} \times \mathbf{c}|}

Option A,

\frac{[a \mathbf{b} \mathbf{c}]}{|\mathbf{a} \times \mathbf{b}+\mathbf{c} \times \mathbf{a}+\mathbf{b} \times \mathbf{c}|}

, is correct because it is the formula for the length of the perpendicular from the origin to the plane passing through three non-collinear points a, b, and c.

28 Medium

28 of 38

Application +4 | -1

{a}, {b}, {c}

are non-coplanar vectors and

{d}

is a unit vector, then the value of

|({a} \cdot {d})({b} \times {c})+({b} \cdot {d})({c} \times {a})+({c} \cdot {d})({a} \times {b})|

View answer and explanation

Correct option: C

1[{a} {b} {c}]

Option C (

1[{a} {b} {c}]

) is correct because the expression simplifies to the scalar triple product of the vectors

{a}, {b}, {c}

29 Medium

29 of 38

Application +4 | -1

If three conterminous edges of a parallelopiped are represented by

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

and

\mathbf{c}-\mathbf{a}

, then its volume is

View answer and explanation

Correct option: D

Option D (0) is correct because the volume of a parallelepiped formed by vectors a-b, b-c, and c-a is always zero, as they are coplanar.

30 Medium

30 of 38

Analyzing +4 | -1

Let

O

be the origin and the position vector of the point

P

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

. If the position vectors of the points

A, B

and

C

are

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

and

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

respectively, then the projection of the vector

\overrightarrow{O P}

on a vector perpendicular to the vectors

\overrightarrow{A B}

and

\overrightarrow{A C}

View answer and explanation

Correct option: A

Option A (3) is correct because the projection of OP on a vector perpendicular to AB and AC is calculated using the scalar triple product.

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