If ${\bf{\vec a}},\,\,{\bf{\vec b}},\,\,{\bf{\vec c}}$ are non-coplanar vectors, then $\left| {\,\begin{array}{*{20}{c}} {{\bf{\vec a}} \cdot {\bf{\vec a}}}&{{\bf{\vec a}} \cdot {\bf{\vec b}}}&{{\bf{\vec a}} \cdot {\bf{\vec c}}}\\ {{\bf{\vec b}} \cdot {\bf{\vec a}}}&{{\bf{\vec b}} \cdot {\bf{\vec b}}}&{{\bf{\vec b}} \cdot {\bf{\vec c}}}\\ {{\bf{\vec c}} \cdot {\bf{\vec a}}}&{{\bf{\vec c}} \cdot {\bf{\vec b}}}&{{\bf{\vec c}} \cdot {\bf{\vec c}}} \end{array}\,} \right|$ is equal to

${\left[ {\;{\bf{\vec a}}\;\,{\bf{\vec b}}\;\,{\bf{\vec c}}\;} \right]^2}$

$\left[ {\;{\bf{\vec a}}\;\,{\bf{\vec b}}\;\,{\bf{\vec c}}\;} \right]$

${\left[ {\;{\bf{\vec a}}\,\;{\bf{\vec b}}\;\,{\bf{\vec c}}\;} \right]^{1/3}}$

The equation of the plane passing through three non-collinear points ${\bf{\vec a}},{\bf{\vec b}},{\bf{\vec c}}$ is

${\bf{\vec r}}\cdot\left( {{\bf{\vec b}} \times {\bf{\vec c}} + {\bf{\vec c}} \times {\bf{\vec a}} + {\bf{\vec a}} \times {\bf{\vec b}}} \right) = \left[ {{\bf{\vec a}}\;{\bf{\vec b}}\;{\bf{\vec c}}} \right]$

${\bf{\vec r}}\cdot\left( {{\bf{\vec b}} \times {\bf{\vec c}} + {\bf{\vec c}} \times {\bf{\vec a}} + {\bf{\vec a}} \times {\bf{\vec b}}} \right) = 0$

${\bf{\vec r}}\cdot\left( {{\bf{\vec a}} \times \left( {{\bf{\vec b}} \times {\bf{\vec c}}} \right)} \right) = \left[ {{\bf{\vec a}}\;{\bf{\vec b}}\;{\bf{\vec c}}} \right]$

${\bf{\vec r}}\cdot\left( {{\bf{\vec a}} + {\bf{\vec b}} + {\bf{\vec c}}} \right) = 0$

$\left( {{\bf{\vec a}} - {\bf{\vec b}}} \right)\cdot\left\{ {\left( {{\bf{\vec b}} - {\bf{\vec c}}} \right) \times \left( {{\bf{\vec c}} - {\bf{\vec a}}\;} \right)} \right\}$ is equal to

$2\;{\bf{\vec a}}\cdot{\bf{\vec b}} \times {\bf{\vec c}}$

${\bf{\vec a}}\cdot{\bf{\vec b}} \times \;{\bf{\vec c}}$

${\bf{\vec a}}\cdot\;{\bf{\vec b}}\;$

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JEE • MATHEMATICS • SCALAR TRIPLE PRODUCT

Scalar triple product practice · Mixed difficulty

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Showing 181-210 of 230 questions

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181 Medium

181 of 230

Understanding +4 | -1

If a, b and

\mathbf{c}

be any three non-zero and non-coplanar vectors, then any vector

r

is equal to
where,

x=\frac{[\mathbf{r} \mathbf{b} \mathbf{c}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}, y=\frac{[\mathbf{r} \mathbf{c} \mathbf{a}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]^{,}} z=\frac{[\mathbf{r} \mathbf{a} \mathbf{b}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}

View answer and explanation

Correct option: B

x a+y b+x

Option D (None of these) is incorrect, the correct answer should be

\mathbf{r} = x\mathbf{a} + y\mathbf{b} + z\mathbf{c}

, but it is not listed in the given options. The correct answer is a linear combination of the vectors a, b, and c.

182 Easy

182 of 230

Application +4 | -1

Let

\mathbf{\vec v} = 2\mathbf{\hat i} + \mathbf{\hat j} - \mathbf{\hat k}\mathrm{\;and\;}\mathbf{\vec w} = \mathbf{\hat i} + 3\mathbf{\hat k},\mathrm{\;If\;}\mathbf{\vec u}

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

\left[ {\;\mathbf{\vec u}\;\mathbf{\vec v}\;\mathbf{\vec w}\mathrm{\;}} \right]

View answer and explanation

Correct option: C

\sqrt {59}

Option C (

\sqrt {59}

) is correct because the maximum value of the scalar triple product is the magnitude of the vector resulting from the cross product of

\vec v

and

\vec w

, since

\vec u

is a unit vector.

183 Easy

183 of 230

Understanding +4 | -1

If a, b, c and p, q, r are reciprocal system of vectors, then

a \times p + b \times q + c \times r

is equal to

View answer and explanation

Correct option: C

Option C (0) is correct because the expression simplifies to zero using the properties of reciprocal systems of vectors.

184 Easy

184 of 230

Understanding +4 | -1

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

are non-coplanar vectors, then

\left| {\,\begin{array}{cccccccccccccccccccc} {\mathbf{\vec a} \cdot \mathbf{\vec a}}&{\mathbf{\vec a} \cdot \mathbf{\vec b}}&{\mathbf{\vec a} \cdot \mathbf{\vec c}}\\ {\mathbf{\vec b} \cdot \mathbf{\vec a}}&{\mathbf{\vec b} \cdot \mathbf{\vec b}}&{\mathbf{\vec b} \cdot \mathbf{\vec c}}\\ {\mathbf{\vec c} \cdot \mathbf{\vec a}}&{\mathbf{\vec c} \cdot \mathbf{\vec b}}&{\mathbf{\vec c} \cdot \mathbf{\vec c}} \end{array}\,} \right|

is equal to

View answer and explanation

Correct option: A

{\left[ {\;\mathbf{\vec a}\;\,\mathbf{\vec b}\;\,\mathbf{\vec c}\;} \right]^2}

Option A (

{\left[ {\;\mathbf{\vec a}\;\,\mathbf{\vec b}\;\,\mathbf{\vec c}\;}} \right]^2}

) is correct because the determinant of the dot products is equivalent to the square of the scalar triple product.

185 Medium

185 of 230

Application +4 | -1

The distance between the lines

\bar{r}=(\bar{i}+2 \bar{j}-4 \bar{k})+\lambda(2 \bar{i}+3 \bar{j}+6 \bar{k})

\bar{r}=(3 \bar{i}+3 \bar{j}-5 \bar{k})+\mu(2 \bar{i}+3 \bar{j}+6 \bar{k})

View answer and explanation

Correct option: C

\frac{\sqrt{293}}{7}

Option B (

\frac{{\sqrt{{293}}}}{{7}}

) is correct because it represents the distance between the two lines, calculated using the formula for the distance between parallel lines in vector form.

186 Medium

186 of 230

Application +4 | -1

Given vectors

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

such that

\mathbf{a}(\mathbf{b} \times \mathbf{c})=\lambda \neq 0

, the value of

(\mathbf{b} \times \mathbf{c}) .(\mathbf{a}+\mathbf{b}+\mathbf{c}) / \lambda

View answer and explanation

Correct option: C

-3 \lambda

Option B (1) is correct because

(\mathbf{b} \times \mathbf{c}) .(\mathbf{a}+\mathbf{b}+\mathbf{c}) / \lambda

simplifies to

\lambda / \lambda = 1

since

(\mathbf{b} \times \mathbf{c}) . \mathbf{b}

and

(\mathbf{b} \times \mathbf{c}) . \mathbf{c}

are zero.

187 Easy

187 of 230

Application +4 | -1

\vec{u}=\hat{j}+4 \hat{k}, \vec{v}=\hat{i}+3 \hat{k}

and

\vec{w}=\cos \theta \hat{i}+\sin \theta \hat{j}

are vectors in 3-dimensional space, then the maximum possible value of

|\vec{u} \times \vec{v} \cdot \vec{w}|

View answer and explanation

Correct option: B

Option B (5) is correct because the maximum value of the scalar triple product is found by maximizing the expression -3cos(θ) + 4sin(θ).

188 Easy

188 of 230

Application +4 | -1

If the shortest distance between the lines

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

and

\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}

\frac{13}{\sqrt{29}},

then a value of

\lambda

is:

View answer and explanation

Correct option: C

Option C (1) is correct because it satisfies the condition for the shortest distance between the two given lines being equal to

\frac{13}{\sqrt{29}}

189 Easy

189 of 230

Application +4 | -1

Let

\mathbf{\vec u} = \widehat \mathbf{i} + \widehat \mathbf{j},\;\mathbf{\vec v} = \;\widehat \mathbf{i} - \widehat \mathbf{j}

and

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

. If

{\mathbf{\hat n}}

is a unit vector such that

\mathbf{\vec u}.\mathbf{\hat n} = 0

and

\mathbf{\vec v}.\mathbf{\hat n} = 0

, then

\left| {\vec w.\mathbf{\hat n}} \right|

is equal to

View answer and explanation

Correct option: D

The correct answer is 3 because the unit vector

{\mathbf{\hat n}}

is perpendicular to both

\mathbf{\vec u}

and

\mathbf{\vec v}

, and

\left| {\vec w.{\mathbf{\hat n}}} \right|

is calculated using the cross product.

190 Easy

190 of 230

Understanding +4 | -1

The equation of the plane passing through three non-collinear points

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

View answer and explanation

Correct option: B

\mathbf{\vec r}\cdot\left( {\mathbf{\vec b} \times \mathbf{\vec c} + \mathbf{\vec c} \times \mathbf{\vec a} + \mathbf{\vec a} \times \mathbf{\vec b}} \right) = \left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]

Option B (

\mathbf{\vec r}\cdot\left( {\mathbf{\vec b} \times \mathbf{\vec c} + \mathbf{\vec c} \times \mathbf{\vec a} + \mathbf{\vec a} \times \mathbf{\vec b}} \right) = \left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]

) is correct because it is the standard formula for the equation of a plane passing through three non-collinear points in vector form.

191 Easy

191 of 230

Application +4 | -1

The shortest distance between the lines

\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}

and

\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}

View answer and explanation

Correct option: B

6 \sqrt{3}

Option B (

6 \sqrt{3}

) is correct because it is the shortest distance between the two lines calculated using the formula involving the cross product of direction vectors and the difference of position vectors.

192 Easy

192 of 230

Application +4 | -1

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

are non-coplanar non-null vetors stch that

\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=2

, then

\left\{\left[\begin{array}{llll}\vec{a} \times \vec{b} & \vec{b} \times \vec{c} & \vec{c} \times \vec{a}\end{array}\right]\right\}^2=

View answer and explanation

Correct option: B

Option B (16) is correct because

\left[\begin{array}{lll}\vec{a} \times \vec{b} & \vec{b} \times \vec{c} & \vec{c} \times \vec{a}\end{array}\right] = \left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]^2

, and thus the square of that is

\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]^4 = 2^4 = 16

193 Easy

193 of 230

Analyzing +4 | -1

If the vectors

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

are non-coplanar und

l, m, n

are distinct scalars such that

[l \vec{a}+m \vec{b}+n \vec{c} \,\,l \vec{b}+m \vec{c}+n \vec{a} \,\,l \vec{c}+m \vec{a}+n \vec{b}]=0

then

View answer and explanation

Correct option: B

l+m+n=0

Option B (

l+m+n=0

) is correct because the determinant equation simplifies to

(l+m+n)(l^2+m^2+n^2-l m-m n-n l)=0

, and since

l, m, n

are distinct,

l+m+n=0

194 Easy

194 of 230

Application +4 | -1

Let

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

be three non-coplanar vectors and let

\mathbf{\vec p},\,\,\mathbf{\vec q}\mathrm{\;}\,\,\,\mathrm{and}\,\mathrm{\;}\mathbf{\vec r}

be vector defined by the relations.

\mathbf{\vec p} = \frac{{\mathbf{\vec b} \times \mathbf{\vec c}}}{{\left[ {\;\mathbf{\vec a}\,\;\mathbf{\vec b}\,\;\mathbf{\vec c}\;} \right]}},\,\,\mathbf{\vec q} = \frac{{\mathbf{\vec c} \times \mathbf{\vec a}}}{{\left[ {\mathbf{\vec a}\;\,\mathbf{\vec b}\;\,\mathbf{\vec c}\;} \right]}}\,\,\mathrm{and\;}\,\,\mathbf{\vec r} = \frac{{\mathbf{\vec a} \times \mathbf{\vec b}}}{{\left[ {\mathbf{\vec a}\;\,\mathbf{\vec b}\;\,\mathbf{\vec c}\;} \right]}}

. Then, the value of the expression

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

is equal to

View answer and explanation

Correct option: D

Option D (3) is correct because by using the properties of reciprocal vectors, we can simplify the expression to equal 3.

195 Easy

195 of 230

Application +4 | -1

Let

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

be three unit vectors such that

\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}=0.

If the angle between

\vec{b}

and

\vec{c}

\frac{\pi}{3},

then the volume of the parallelopiped whose three coterminous edges are

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

View answer and explanation

Correct option: A

\frac{\sqrt{3}}{2}

cubic units

Option A (

\frac{\sqrt{3}}{2}

cubic units) is correct because the volume of the parallelepiped is calculated using the scalar triple product, considering the given dot product and unit vector conditions.

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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 Easy

196 of 230

Analyzing +4 | -1

\vec{u}, \vec{v}, \vec{w}

are three non-coplanar vectors, then

(\vec{u}+\vec{v}-\vec{w}) \cdot(\vec{u}-\vec{v}) \times(\vec{v}-\vec{w})

equals

View answer and explanation

Correct option: A

\vec{u} \cdot(\vec{v} \times \vec{w})

Option A (

\vec{u} \cdot(\vec{v} \times \vec{w})

) is correct because the triple scalar product simplifies to

\vec{u} \cdot(\vec{v} \times \vec{w})

197 Easy

197 of 230

Application +4 | -1

The volume (in cubic unit) of the tetrahedron with edges

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

and

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

View answer and explanation

Correct option: B

\frac{2}{3}

Option B (

\frac{2}{3}

) is correct because it is the volume of the tetrahedron calculated using the scalar triple product of the edge vectors.

198 Medium

198 of 230

Application +4 | -1

The line

I_1

passes through the point

(2,6,2)

and is perpendicular to the plane

2 x+y-2 z=10

. Then the shortest distance between the line

h

and the line

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

View answer and explanation

Correct option: C

Option C (9) is correct because it represents the shortest distance between the two lines, calculated using the formula for skew lines.

199 Medium

199 of 230

Application +4 | -1

\mathbf{a}=(1,-1,1)

and

\mathbf{c}=(-1,-1,0)

, then the vector

\mathbf{b}

satisfying

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

and

\mathbf{a} \cdot \mathbf{b}=1

View answer and explanation

Correct option: B

(0,0,1)

Option B ((0,0,1)) is correct because it satisfies both the cross product and dot product conditions with the given vectors a and c.

200 Easy

200 of 230

Application +4 | -1

a=-3 i+7 j+5 k, \quad \mathbf{b}=-3 i+7 j-3 k, \quad c=7 i-5 j-3 k

are the three coterminous edges of a parallelopiped, then

View answer and explanation

Correct option: C

272

Option C (272) is correct because the volume of the parallelopiped is calculated by the absolute value of the determinant of the matrix formed by the vectors, resulting in 272.

201 Easy

201 of 230

Analyzing +4 | -1

\left( {\mathbf{\vec a} - \mathbf{\vec b}} \right)\cdot\left\{ {\left( {\mathbf{\vec b} - \mathbf{\vec c}} \right) \times \left( {\mathbf{\vec c} - \mathbf{\vec a}\;} \right)} \right\}

is equal to

View answer and explanation

Correct option: C

Option C (0) is correct because the scalar triple product of linearly dependent vectors is zero.

202 Easy

202 of 230

Understanding +4 | -1

The equation

{\mathbf{r}^{\vec 2}} - 2\mathrm{\;}\mathbf{\vec r}\cdot\mathbf{\vec c} + h = 0,\left| {\mathbf{\vec c}} \right| > \sqrt h

, represent

View answer and explanation

Correct option: D

Sphere

Option D (Sphere) is correct because the given equation is a form of the sphere equation in vector form.

203 Easy

203 of 230

Analyzing +4 | -1

Let a, b, c are three non-coplanar vectors and d be a non-zero vector, which is perpendicular to

a+b+c

Now, If

d = (\sin x)(a \times b) + (\cos y)(b \times c) + 2(c \times a)

then the minimum value of

\left( {{x^2} + {y^2}} \right)

View answer and explanation

Correct option: D

\frac{{5{\pi ^2}}}{4}

The correct answer is D (

\frac{{5{\pi ^2}}}{{4}}

) because it is obtained by finding the values of x and y that satisfy the given conditions and then calculating the minimum value of x² + y².

204 Easy

204 of 230

Analyzing +4 | -1

If the foot of the perpendicular from point

(4,3,8)

on the line

L_1: \frac{x-a}{2}=\frac{y-2}{m}=\frac{z-b}{4}, m \neq 0

(3,5,7)

then the shortest distance between the line

L_1

and the line

L_2 \Rightarrow \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}

is equal to

View answer and explanation

Correct option: B

\frac{1}{\sqrt{6}}

Option B (

\frac{1}{\sqrt{6}}

) is correct because it represents the shortest distance between the two lines, calculated using the formula involving the vector difference and cross product of direction vectors.

205 Easy

205 of 230

Analyzing +4 | -1

If V is the volume of the parallelopiped having three coterminous edges as a, b and c, then the volume of parallelopiped having three coterminous edges as

\alpha = (a \cdot a)a + (a \cdot b)b + (a \cdot c)c,\,\,\,\,\,\beta = (a \cdot b)a + (b \cdot b)b + (b \cdot c)c\,\,

and

\gamma = (a \cdot c)a + (b \cdot c)b + (c \cdot c)c

View answer and explanation

Correct option: A

{V^3}

Option A (

{V^3}

) is correct because the volume of the new parallelepiped is the cube of the original volume.

206 Easy

206 of 230

Application +4 | -1

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

and

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

, then a.

(b \times c)=

View answer and explanation

Correct option: C

Option C (12) is correct because the scalar triple product a.(b x c) is calculated correctly using the determinant method.

207 Easy

207 of 230

Application +4 | -1

Given unit vectors

\mathbf{m}, \mathbf{n}

and

\mathbf{p}

such that angle between

\mathbf{m}

and

\mathbf{n}=

Angle between

\mathbf{p}

and

(\mathbf{m} \times \mathbf{n})=\frac{\pi}{6}

, then [npm] is equal to

View answer and explanation

Correct option: A

\sqrt{3} / 4

Option A (

\sqrt{3} / 4

) is correct because it is the result of calculating the scalar triple product [npm] using the given angles between the vectors.

208 Easy

208 of 230

Application +4 | -1

The volume of the parallelopiped whose edges are represented by

-12 i+a k, 3 j-k

and

2 i+j-15 k

is 546 . Then

\alpha=

View answer and explanation

Correct option: C

-3

Option C (-3) is correct because the volume of the parallelepiped is given by the absolute value of the scalar triple product of the vectors representing its edges, and solving for alpha gives -3.

209 Easy

209 of 230

Understanding +4 | -1

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

are three non-coplanar vectors, that

\frac{\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}}{\mathbf{c} \times \mathbf{a} \cdot \mathbf{b}}+\frac{\mathbf{b} \cdot \mathbf{a} \times \mathbf{c}}{\mathbf{c} \cdot \mathbf{a} \times \mathbf{b}}

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the expression simplifies to

\frac{{[abc]}}{{[abc]}} - \frac{{[abc]}}{{[abc]}} = 1 - 1 = 0

210 Easy

210 of 230

Application +4 | -1

The value of '

a

' so that the volume of the parallelopiped formed by

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

and

a \hat{i}+\hat{k}

becomes minimum, is

View answer and explanation

Correct option: C

\frac{1}{\sqrt{3}}

Option C (

\frac{{1}}{{\sqrt{{3}}}}

) is correct because it minimizes the volume of the parallelepiped, which is found by taking the derivative of the volume with respect to 'a' and setting it to zero.

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