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JEE • MATHEMATICS • SCALAR TRIPLE PRODUCT

Scalar triple product practice · Mixed difficulty

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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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121 Easy
121 of 230
Application +4 | -1
The volume of the tetrahedron whose vertices are the points i^,i^+j^,i^+j^+k^\hat{i}, \hat{i}+\hat{j}, \hat{i}+\hat{j}+\hat{k} and 2i^+3j^+λk^2 \hat{i}+3 \hat{j}+\lambda \hat{k} is 1/61 / 6 units. Then, the values of λ\lambda
View answer and explanation
Correct option: D
is any real value
Option D (is any real value) is correct because the volume of the tetrahedron is independent of λ\lambda, as the scalar triple product simplifies to a constant value regardless of λ\lambda.
122 Easy
122 of 230
Analyzing +4 | -1
If a=110(3i^+k^)\mathbf{a}=\frac{1}{\sqrt{10}}(3 \hat{\mathbf{i}}+\hat{\mathbf{k}}) and b=17(2i^+3j^−6k^)\mathbf{b}=\frac{1}{7}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}), then the value of (2−b)⋅[(a×b)×(a+2b)](2-b) \cdot[(\mathbf{a} \times \mathbf{b}) \times(\mathbf{a}+2 \mathbf{b})] is
View answer and explanation
Correct option: D
-5
Option D (-5) is correct because by applying vector triple product properties and simplifying, the expression evaluates to -5.
123 Easy
123 of 230
Application +4 | -1
If a⃗{\mathbf{\vec a}} and b⃗{\mathbf{\vec b}} are unit vectors and ∣a⃗+b⃗∣=1\left| {\mathbf{\vec a} + \mathbf{\vec b}} \right| = 1, then ∣a⃗+b⃗∣\left| {\mathbf{\vec a} + \mathbf{\vec b}} \right| is equal to
View answer and explanation
Correct option: D
3\sqrt 3
Option D (3\sqrt 3) is correct because using the given conditions and the properties of dot products, the magnitude of the difference vector is found to be 3\sqrt{3}.
124 Easy
124 of 230
Application +4 | -1
A parallelopiped is formed by planes drawn parallel to coordinate axes through the points A=(1,2,3)A=(1,2,3) and B=(9,8,5)B=(9,8,5). The volume of that parallelopiped is equal to (in cubic wits)
View answer and explanation
Correct option: D
96
Option D (96) is correct because it represents the volume of the parallelepiped formed by the planes drawn parallel to the coordinate axes through the given points.
125 Easy
125 of 230
Analyzing +4 | -1
If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are any three vectors such that (a⃗+b⃗)⋅c⃗=(a⃗−b⃗)⋅c⃗=0,(\vec{a}+\vec{b}) \cdot \vec{c}=(\vec{a}-\vec{b}) \cdot \vec{c}=0, then (a⃗×b⃗)×c⃗(\vec{a} \times \vec{b}) \times \vec{c} is
View answer and explanation
Correct option: A
0→\overrightarrow{0}
Option A (0→\overrightarrow{0}) is correct because the given conditions imply that both a⃗\vec{a} and b⃗\vec{b} are orthogonal to c⃗\vec{c}, leading to the result (a⃗×b⃗)×c⃗=0→(\vec{a} \times \vec{b}) \times \vec{c} = \overrightarrow{0}.
126 Easy
126 of 230
Application +4 | -1
If r⋅i=r⋅j=r⋅k\mathbf{r} \cdot \mathbf{i}=\mathbf{r} \cdot \mathbf{j}=\mathbf{r} \cdot \mathbf{k} and ∣r∣=3|\mathbf{r}|=3, then r=\mathbf{r}=
View answer and explanation
Correct option: D
±3(i+j+k)\pm \sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k})
Option D (±3(i+j+k)\pm \sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k})) is correct because the given conditions imply that all components of r are equal, and the magnitude condition determines their value.
127 Easy
127 of 230
Analyzing +4 | -1
Let the vectors
(2+a+b)i^+(a+2b+c)j^−(b+c)k^(2 + a + b)\widehat i + (a + 2b + c)\widehat j - (b + c)\widehat k,
(1+b)i^+2bj^−bk^(1 + b)\widehat i + 2b\widehat j - b\widehat k and (2+b)i^+2bj^+(1−b)k^(2 + b)\widehat i + 2b\widehat j + (1 - b)\widehat k, a,b,c∈Ra,b,c \in R be co-planar.
Then, which of the following is true?
View answer and explanation
Correct option: A
2b=a+c2b = a + c
Option A (2b=a+c2b = a + c) is correct because the coplanarity condition leads to a determinant equation that simplifies to this relationship.
128 Medium
128 of 230
Application +4 | -1
The equation ∣r∣2−r.(2i+4j−2k)−10=0|\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.
129 Medium
129 of 230
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
[abc]∣a×b+c×a+b×c∣\frac{[a \mathbf{b} \mathbf{c}]}{|\mathbf{a} \times \mathbf{b}+\mathbf{c} \times \mathbf{a}+\mathbf{b} \times \mathbf{c}|}
Option A, [abc]∣a×b+c×a+b×c∣\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.
130 Medium
130 of 230
Application +4 | -1
If a,b,c{a}, {b}, {c} are non-coplanar vectors and d{d} is a unit vector, then the value of ∣(a⋅d)(b×c)+(b⋅d)(c×a)+(c⋅d)(a×b)∣|({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[abc]1[{a} {b} {c}]
Option C (1[abc]1[{a} {b} {c}]) is correct because the expression simplifies to the scalar triple product of the vectors a,b,c{a}, {b}, {c}.
131 Hard
131 of 230
Application +4 | -1
A unit vector coplanar with \widehat \mathbf{i} + \widehat \mathbf{j} + 2\mathbf{\hat k}\widehat \mathbf{i} + \widehat \mathbf{j} + 2\mathbf{\hat k} and \widehat \mathbf{i} + 2\widehat \mathbf{j} + \mathbf{\hat k}\widehat \mathbf{i} + 2\widehat \mathbf{j} + \mathbf{\hat k}, and perpendicular to \widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k} is
View answer and explanation
Correct option: A
\left( {\frac{{\widehat \mathbf{j} - \mathbf{\hat k}}}{{\sqrt 2 }}} \right)\left( {\frac{{\widehat \mathbf{j} - \mathbf{\hat k}}}{{\sqrt 2 }}} \right)
Option A (\left( {\frac{{\widehat \mathbf{j} - \mathbf{\hat k}}}{{\sqrt 2 }}} \right)\left( {\frac{{\widehat \mathbf{j} - \mathbf{\hat k}}}{{\sqrt 2 }}} \right)) is correct because it is a unit vector that is coplanar with the given vectors and perpendicular to \widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}.
132 Easy
132 of 230
Application +4 | -1
Let a⃗,  b⃗  and  c⃗\mathbf{\vec a},\;\mathbf{\vec b}\;\mathrm{and\;}\mathbf{\vec c} be three non-coplanar vectors, and let p⃗,  q⃗  and  r⃗\mathbf{\vec p},\;\mathbf{\vec q}\;\mathrm{and\;}\mathbf{\vec r} be vectors defined by
the relations

p⃗=b⃗×c⃗[a⃗  b⃗  c⃗]  ,  q⃗=c⃗×a⃗[a⃗  b⃗  c⃗]and    r⃗=a⃗×b⃗[a⃗  b⃗  c⃗]    \mathbf{\vec p} = \frac{{\mathbf{\vec b} \times \mathbf{\vec c}}}{{\left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]}}\mathrm{\;},\mathrm{\;}\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]}}\mathrm{\;\;}

Then, the value of the expression

(a⃗+b⃗)⋅p⃗+(b⃗+  c⃗)⋅q⃗+(c⃗+a⃗)⋅r⃗\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
3
Option D (3) is correct because it is the result of simplifying the expression using properties of scalar triple products and reciprocal vectors.
133 Easy
133 of 230
Analyzing +4 | -1
Unit vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are coplanar. A unit vector d⃗\vec{d} is perpendicular to them. If (a⃗×b⃗)×(c⃗×d⃗)=16i^−13j^+13k^(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\frac{1}{6} \hat{i}-\frac{1}{3} \hat{j}+\frac{1}{3} \hat{k} and the angle between a⃗\vec{a} and b⃗\vec{b} is 30∘30^{\circ}, then c⃗\vec{c} is/are
View answer and explanation
Correct option: D
±13(−i^+2j^−2k^)\pm \frac{1}{3}(-\hat{i}+2 \hat{j}-2 \hat{k})
Option D (±13(−i^+2j^−2k^)\pm \frac{1}{3}(-\hat{i}+2 \hat{j}-2 \hat{k})) is correct because it represents the possible directions of vector c⃗\vec{c} based on the given conditions and vector triple product.
134 Easy
134 of 230
Application +4 | -1
If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are three non-coplanar vectors, then a vector r⃗\vec{r} satisfying r⃗⋅a⃗=r⃗⋅b⃗=r⃗⋅c⃗=1\vec{r} \cdot \vec{a}=\vec{r} \cdot \vec{b}=\vec{r} \cdot \vec{c}=1, is
View answer and explanation
Correct option: B
1[a⃗b⃗c⃗]{a⃗×b⃗+b⃗×c⃗+c⃗×a⃗}\frac{1}{\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]}\{\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}\}
Option B (1[a⃗b⃗c⃗]{a⃗×b⃗+b⃗×c⃗+c⃗×a⃗}\frac{1}{\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]}\{\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}\}) is correct because it satisfies the conditions r⃗⋅a⃗=r⃗⋅b⃗=r⃗⋅c⃗=1\vec{r} \cdot \vec{a}=\vec{r} \cdot \vec{b}=\vec{r} \cdot \vec{c}=1.
135 Medium
135 of 230
Application +4 | -1
If three conterminous edges of a parallelopiped are represented by a−b,b−c\mathbf{a}-\mathbf{b}, \mathbf{b}-\mathbf{c} and c−a\mathbf{c}-\mathbf{a}, then its volume is
View answer and explanation
Correct option: D
0
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.
136 Easy
136 of 230
Understanding +4 | -1
Statement 1: The vectors a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c} lie in the same plane if and only if a⃗⋅(b⃗×c⃗)=0\vec{a} \cdot(\vec{b} \times \vec{c})=0

Statement 2: The vectors u⃗\vec{u} and v⃗\vec{v} are perpendicular if and only if u⃗⋅v⃗=0\vec{u} \cdot \vec{v}=0 where u⃗×v⃗\vec{u} \times \vec{v} is a vector perpendicular to the plane of u⃗\vec{u} and v⃗\vec{v} .
View answer and explanation
Correct option: C
Statement 1 is true, Statement 2 is false.
Statement 1 is true because it describes the condition for coplanarity of three vectors. Statement 2 is false because it states that u⃗×v⃗\vec{u} \times \vec{v} is perpendicular to the plane, but it is actually u⃗⋅v⃗=0\vec{u} \cdot \vec{v} = 0 that indicates perpendicularity.
137 Easy
137 of 230
Application +4 | -1
If a⃗=2i^−3j^+5k^\mathbf{\vec a} = 2\mathbf{\hat i} - 3\mathbf{\hat j} + 5\mathbf{\hat k}, b⃗=3i^−4j^+5k^\mathbf{\vec b} = 3\mathbf{\hat i} - 4\mathbf{\hat j} + 5\mathbf{\hat k} and c⃗=5i^−3j^−2k^\mathbf{\vec c} = 5\mathbf{\hat i} - 3\mathbf{\hat j} - 2\mathbf{\hat k}, then the volume of the parallelopiped with coterminous edges a⃗+b⃗,b⃗+c⃗,c⃗+a⃗\mathbf{\vec a} + \mathbf{\vec b},\mathbf{\vec b} + \mathbf{\vec c},\mathbf{\vec c} + \mathbf{\vec a} is
View answer and explanation
Correct option: D
8
Option D (8) is correct because it is the volume of the parallelepiped calculated using the scalar triple product of the given vectors.
138 Easy
138 of 230
Application +4 | -1
If a⃗=i^+2j^+k^,b⃗=3(i^−j^+k^),a⃗⋅c⃗=3\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3(\hat{i}-\hat{j}+\hat{k}), \vec{a} \cdot \vec{c}=3
a⃗×c⃗=b⃗\vec{a} \times \vec{c}=\vec{b} then a⃗⋅[(c⃗×b⃗)−b⃗−c⃗]=\vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}]=
View answer and explanation
Correct option: A
24
Option A (24) is correct because by using the vector triple product and scalar triple product properties, the expression simplifies to 24.
139 Easy
139 of 230
Application +4 | -1
If ∣a∣=3,∣b∣=4,∣c∣=5|\mathbf{a}|=3,|\mathbf{b}|=4,|\mathbf{c}|=5 and a+b+c=0\mathbf{a}+\mathbf{b}+\mathbf{c}=0, then the angle between a and bb is
View answer and explanation
Correct option: D
π2\frac{\pi}{2}
Option D (π2\frac{\pi}{2}) is correct because the angle between vectors a and b is found to be 90 degrees using the given magnitudes and the condition a + b + c = 0.
140 Easy
140 of 230
Analyzing +4 | -1
The vectors a⃗=xi^+(x+1)j^+(x+2)k^, \vec a = x\hat i + (x + 1)\hat j + (x + 2)\hat k{\text{, }} b⃗=(x+3)i^+(x+4)j^+(x+5)k^\vec{b}=(x+3) \hat{i}+(x+4) \hat{j}+(x+5) \hat{k}
and, c⃗=(x+6)i^+(x+7)j^+(x+8)k^\vec{c}=(x+6) \hat{i}+(x+7) \hat{j}+(x+8) \hat{k} are coplanar for
View answer and explanation
Correct option: A
all values of xx
Option A (all values of x) is correct because the scalar triple product of the given vectors is always zero, indicating that they are coplanar for any value of x.
141 Easy
141 of 230
Application +4 | -1
If \widehat \mathbf{i} - \mathbf{\hat k},\;\mathrm{\lambda \;}\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - \lambda } \right)\mathbf{\hat k}\widehat \mathbf{i} - \mathbf{\hat k},\;\mathrm{\lambda \;}\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - \lambda } \right)\mathbf{\hat k} and \mathrm{\mu }\;\widehat \mathbf{i} + \mathrm{\lambda \;}\widehat \mathbf{j} + \left( {1 + \mathrm{\lambda \;}\widehat \mathbf{j} - \mathrm{\mu }} \right)\mathbf{\hat k}\mathrm{\mu }\;\widehat \mathbf{i} + \mathrm{\lambda \;}\widehat \mathbf{j} + \left( {1 + \mathrm{\lambda \;}\widehat \mathbf{j} - \mathrm{\mu }} \right)\mathbf{\hat k} are three coterminal edges of a parallelopiped, then its volume depends on
View answer and explanation
Correct option: D
Neither λ nor μ
Option D is correct because the volume of the parallelepiped is independent of both λ and μ, and is equal to 1.
142 Easy
142 of 230
Application +4 | -1
The value of λ\lambda, for which the four points 2 i^+3 j^−k^,   i^+2 j^+3 k^,  3 i^+4 j^−2 k^,   i^−λ j^+6 k^2\,\mathbf{\hat i} + 3\,\mathbf{\hat j} - \mathbf{\hat k},\,\;\mathbf{\hat i} + 2\,\mathbf{\hat j} + 3\,\mathbf{\hat k},\mathrm{\;}3\,\mathbf{\hat i} + 4\,\mathbf{\hat j} - 2\,\mathbf{\hat k},\mathrm{\;}\,\mathbf{\hat i} - \mathrm{\lambda }\,\mathbf{\hat j} + 6\,\mathbf{\hat k} are coplanar, is

View answer and explanation
Correct option: A
−2- 2
Option A (-2) is correct because the coplanarity condition requires the scalar triple product of the vectors formed by the points to be zero, which leads to an equation that can be solved for λ\lambda.
143 Easy
143 of 230
Application +4 | -1
The distance of the point (−1,2,3)(-1,2,3) from the plane r⃗⋅(i^−2j^+3k^)=10\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=10 parallel to the line of the shortest distance between the lines r⃗=(i^−j^)+λ(2i^+k^)\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k}) and r⃗=(2i^−j^)+μ(i^−j^+k^)\vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k}) is
View answer and explanation
Correct option: C
262 \sqrt{6}
Option C (262 \sqrt{6}) is correct because it is the distance calculated by finding the intersection point of a line parallel to the shortest distance between two given lines and the given plane.
144 Medium
144 of 230
Analyzing +4 | -1
Let OO be the origin and the position vector of the point PP be −i^−2j^+3k-\hat{i}-2 \hat{j}+3 k. If the position vectors of the points A,BA, B and CC are −2i^+j^−3k,2i^+4j^−2k-2 \hat{i}+\hat{j}-3 k, 2 \hat{i}+4 \hat{j}-2 k and −4i^+2j^−k-4 \hat{i}+2 \hat{j}-k respectively, then the projection of the vector OP→\overrightarrow{O P} on a vector perpendicular to the vectors AB→\overrightarrow{A B} and AC→\overrightarrow{A C} is
View answer and explanation
Correct option: A
3
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.
145 Easy
145 of 230
Analyzing +4 | -1
The value of a⃗⋅(b⃗+c⃗)×(a⃗+b⃗+c⃗)\vec{a} \cdot(\vec{b}+\vec{c}) \times(\vec{a}+\vec{b}+\vec{c}), is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the vectors \vec{a}, (\vec{b}+\vec{c}), and (\vec{a}+\vec{b}+\vec{c}) are coplanar, making the scalar triple product equal to zero.
146 Easy
146 of 230
Application +4 | -1
If the volume of a parallelepiped, whose coterminous edges are given by the vectors
a=i^+j^+nk^,    b=2i^+4j^−nk^a = \hat i + \hat j + n\hat k,\,\,\,\,b = 2\hat i + 4\hat j - n\hat k and c=i^+nj^+3k^(n≥0),c = \hat i + n\hat j + 3\hat k\left( {n \ge 0} \right), is 158 cu units, then
View answer and explanation
Correct option: B
b⋅c=10b \cdot c = 10
Option B (b⋅c=10b \cdot c = 10) is correct because after solving for n, the dot product of vectors b and c equals 10.
147 Easy
147 of 230
Application +4 | -1
If the vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} and d⃗\vec{d} are coplanar vectors, then (a⃗×b⃗)×(c⃗×d⃗)(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d}) is equal to
View answer and explanation
Correct option: B
0→\overrightarrow{0}
Option B (0→\overrightarrow{{0}}) is correct because if the vectors are coplanar, then a⃗×b⃗\vec{{a}} \times \vec{{b}} and c⃗×d⃗\vec{{c}} \times \vec{{d}} are parallel, and the cross product of parallel vectors is the zero vector.
148 Easy
148 of 230
Analyzing +4 | -1
Let r, a, b and c be four non-zero vectors such that r.  a=0,r.\,\,a = 0,
∣r×b∣=∣r∣∣b∣,  ∣r×c∣=∣r∥c∣,|r \times b| = |r||b|,\,\,|r \times c| = |r\parallel c\mid , then [a b c] is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the given conditions imply that a is perpendicular to both b and c, making the scalar triple product zero.
149 Easy
149 of 230
Analyzing +4 | -1
If α(a×b)+β(b×c)+γ(c×a)=0\alpha(\mathbf{a} \times \mathbf{b})+\beta(\mathbf{b} \times \mathbf{c})+\gamma(\mathbf{c} \times \mathbf{a})=0, then
View answer and explanation
Correct option: A
a,b,ca, b, c are coplanar if all of α,β,γ≠0\alpha, \beta, \gamma \neq 0
Option A is correct because if all coefficients are non-zero, the scalar triple product must be zero, indicating coplanarity.
150 Medium
150 of 230
Application +4 | -1
If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are vectors such that [a⃗b⃗c⃗]=4[\vec{a} \vec{b} \vec{c}]=4 then [a⃗×b⃗b⃗×c⃗c⃗×a⃗]=[\vec{a} \times \vec{b} \vec{b} \times \vec{c} \vec{c} \times \vec{a}]=
View answer and explanation
Correct option: A
16
Option A (16) is correct because the scalar triple product of cross products simplifies to the square of the original scalar triple product.
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Review SCALAR TRIPLE PRODUCT in your textbook or class notes, then use this set to check your understanding.

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