The vectors $3\hat{i} - \hat{j} + 2\hat{k}$, $2\hat{i} + \hat{j} + 3\hat{k}$ and $\hat{i} + \lambda\hat{j} - \hat{k}$ are coplanar if value of $\lambda$ is (A) $-2$ (B) $0$ (C) $2$ (D) Any real number

The vectors 3hati - hatj + 2hatk, 2hati + hatj + 3hatk — CBSE Class 12 Mathematics 2020 PYQ

Question

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

CBSECBSE Class XII Board 2020MCQ· 1mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Vector Triple Product

The Vector Triple Product

When you cross three vectors together as $\vec{a}\times(\vec{b}\times\vec{c})$ , the result is again a vector — this is the vector triple product. The remarkable thing is that it can always be rewritten without any cross products at all, using only dot products.

The Key Identity (the "BAC $-$ CAB" rule)

\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{a}\cdot\vec{b})\,\vec{c}

A memorable way to recall it: the answer is B times (A dot C) minus C times (A dot B) — "BAC minus CAB". The two survivors are $\vec{b}$ and $\vec{c}$ (the vectors inside the inner bracket); each is scaled by a dot product involving the outside vector $\vec{a}$ .

Why the Result Lies in the Plane of $\vec{b}$ and $\vec{c}$

The inner product $\vec{b}\times\vec{c}$ is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$ . Crossing $\vec{a}$ with that perpendicular swings the result back into the $\vec{b}$ – $\vec{c}$ plane. So the answer must be a combination $\lambda\vec{b}+\mu\vec{c}$ — and the identity tells you exactly what $\lambda$ and $\mu$ are.

Order Matters — the Product Is Not Associative

The brackets are not decoration. In general,

\vec{a}\times(\vec{b}\times\vec{c}) \ne (\vec{a}\times\vec{b})\times\vec{c}.

The left side lies in the plane of $\vec{b},\vec{c}$ ; the right side lies in the plane of $\vec{a},\vec{b}$ . Its own expansion is

(\vec{a}\times\vec{b})\times\vec{c} = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{b}\cdot\vec{c})\,\vec{a},

which is a different vector. Always keep the parentheses where they are given.

Watch out

A frequent slip is to "cancel" and write $\vec{a}\times(\vec{b}\times\vec{c})$ as some multiple of $\vec{a}$ . It is not — the surviving vectors are $\vec{b}$ and $\vec{c}$ , never the outer vector. …

Start your 14-day free trial to unlock the full solution →

Three vectors are coplanar if and only if their scalar triple product is zero; setting the determinant of the matrix formed by their components to zero gives $\lambda = -2$ .

Three vectors are coplanar when they lie in the same plane, which happens precisely when one can be written as a linear combination of the other two. The algebraic test for this is that their scalar triple product vanishes. Geometrically, the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ gives the volume of the parallelepiped spanned by the three vectors — if they're coplanar, that volume collapses to zero.

The scalar triple product can be computed as the determinant of the $3 \times 3$ matrix whose rows (or columns) are the three vectors.

Let me denote:

\vec{a} = 3\hat{i} - \hat{j} + 2\hat{k}, \quad \vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}, \quad \vec{c} = \hat{i} + \lambda\hat{j} - \hat{k}

Set up the determinant condition The vectors are coplanar if and only if

\begin{vmatrix} 3 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & \lambda & -1 \end{vmatrix} = 0

Expand the determinant along the first row

\det = 3 \begin{vmatrix} 1 & 3 \\ \lambda & -1 \end{vmatrix} - (-1) \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} + 2 \begin{vmatrix} 2 & 1 \\ 1 & \lambda \end{vmatrix}

Compute each $2 \times 2$ minor
- First minor: $\begin{vmatrix} 1 & 3 \\ \lambda & -1 \end{vmatrix} = (1)(-1) - (3)(\lambda) = -1 - 3\lambda$
- Second minor: $\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = (2)(-1) - (3)(1) = -2 - 3 = -5$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 21 on this concept.

CBSE 2026Set A1 markMCQ
$\vec{i} \cdot (\vec{j} \times \vec{k}) =$
(a) $1$
(b) $0$
(c) $-1$
(d) $\vec{i}$
›Reveal solutionSolution

$\vec{j}\times\vec{k} = \vec{i}$ , and $\vec{i}\cdot\vec{i} = 1$ .

Using the right-handed rule, $\vec{j} \times \vec{k} = \vec{i}$ . Then

$\vec{i} \cdot (\vec{j} \times \vec{k}) = \vec{i} \cdot \vec{i} = 1.$ …

Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$\vec{a} \cdot (\vec{a} \times \vec{a}) =$
(a) $1$
(b) $0$
(c) $\vec{a}$
(d) $-1$
›Reveal solutionSolution

$\vec{a}\times\vec{a}=\vec{0}$ , hence $\vec{a}\cdot\vec{0}=0$ .

Any vector crossed with itself is the zero vector: $\vec{a}\times\vec{a} = \vec{0}$ . Therefore …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})$
(a) $0$
(b) $-1$
(c) $1$
(d) $3$
›Reveal solutionSolution

$1+(-1)+1=1$ .

$\hat i\cdot(\hat j\times\hat k)=\hat i\cdot\hat i=1$ .

$\hat j\cdot(\hat i\times\hat k)=\hat j\cdot(-\hat j)=-1$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$(\vec{k} \times \vec{j}) \cdot \vec{i} =$
(a) $0$
(b) $1$
(c) $-1$
(d) $2\vec{i}$
›Reveal solutionSolution

This is a scalar triple product; its value is $-1$ .

Using the right-hand cyclic rule, $\vec{j}\times\vec{k}=\vec{i}$ , $\vec{k}\times\vec{i}=\vec{j}$ , $\vec{i}\times\vec{j}=\vec{k}$ . Reversing the order changes the sign, so …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$\hat{i}\cdot(\hat{k}\times\hat{j})-\hat{j}\cdot(\hat{k}\times\hat{i})+\hat{k}\cdot(\hat{i}\times\hat{j})$
(a) $1$
(b) $0$
(c) $-3$
(d) $-1$
›Reveal solutionSolution

Evaluate each triple-scalar-product term using the standard identities $\hat{i}\times\hat{j}=\hat{k}$ , $\hat{j}\times\hat{k}=\hat{i}$ , $\hat{k}\times\hat{i}=\hat{j}$ .

$\hat{k}\times\hat{j}=-(\hat{j}\times\hat{k})=-\hat{i}$ , so $\hat{i}\cdot(\hat{k}\times\hat{j})=\hat{i}\cdot(-\hat{i})=-1$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set D1 markMCQ
$\vec{i} \cdot (\vec{j} \times \vec{k}) =$
(a) $1$
(b) $0$
(c) $-1$
(d) $\vec{i}$
›Reveal solutionSolution

This is the scalar triple product $[\vec{i}\ \vec{j}\ \vec{k}] = 1$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set 65/1/11 markMCQ
$(\hat{i} \times \hat{j}) \cdot \hat{j} + (\hat{j} \times \hat{i}) \cdot \hat{k}$
›Reveal solutionSolution
We evaluate the expression by applying the properties of cross and dot products for orthonormal unit vectors. The first term $(\hat{i} \times \hat{j}) \cdot \hat{j}$ simplifies to $0$ , and the second term $(\hat{j} \times \hat{i}) \cdot \hat{k}$ simplifies to $-1$ . The sum is -1.

The problem asks us to evaluate an expression involving the cross product and dot product of the standard orthonormal unit vectors $\hat{i}$ , $\hat{j}$ , and $\hat{k}$ . These vectors represent the directions along the positive x, y, and z axes, respectively, and each has a magnitude of 1.

The cross product of two vectors results in a vector perpendicular to both original vectors. For $\hat{i}$ , $\hat{j}$ , $\hat{k}$ , they follow a right-hand rule:

$\hat{i} \times \hat{j} = \hat{k}$

$\hat{j} \times \hat{k} = \hat{i}$

$\hat{k} \times \hat{i} = \hat{j}$ The cross product is anti-commutative, meaning reversing the order of the vectors changes the sign of the result: $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$ . For example, $\hat{j} \times \hat{i} = -\hat{k}$ .

The dot product of two vectors results in a scalar. It measures the extent to which two vectors point in the same direction.

If two vectors are orthogonal (perpendicular), their dot product is 0. For example, $\hat{i} \cdot \hat{j} = 0$ .

If two vectors are parallel, their dot product is the product of their magnitudes. For unit vectors, $\hat{a} \cdot \hat{a} = |\hat{a}|^2 = 1^2 = 1$ .

Let's apply these properties to evaluate the given expression term by term.

The expression we need to evaluate is $(\hat{i} \times \hat{j}) \cdot \hat{j} + (\hat{j} \times \hat{i}) \cdot \hat{k}$ .

Evaluate the first term: $(\hat{i} \times \hat{j}) \cdot \hat{j}$

First, we determine the cross product $\hat{i} \times \hat{j}$ .
$\hat{i}$
Substituting this into the first term, we get:

$(\hat{i} \times \hat{j}) \cdot \hat{j} = \hat{k} \cdot \hat{j}$

Next, we evaluate the dot product $\hat{k} \cdot \hat{j}$ . Since $\hat{k}$ and $\hat{j}$ are orthogonal (perpendicular) unit vectors, their dot product is zero.
$\hat{a} \cdot \hat{b} = 0 \quad \text{if } \hat{a} \perp \hat{b}$
Therefore,

$\hat{k} \cdot \hat{j} = 0$

So, the first term evaluates to $0$ .

Tip

This term is a scalar triple product $(\hat{i} \times \hat{j}) \cdot \hat{j}$ . A property of the scalar triple product is that if any two vectors are identical, the value is zero. This is because the three vectors would be coplanar, and the volume of the parallelepiped they form would be zero.

Evaluate the second term: $(\hat{j} \times \hat{i}) \cdot \hat{k}$

First, we determine the cross product $\hat{j} \times \hat{i}$ . The cross product is anti-commutative.
$The anti-commutativity property of the cross product states: \dots$
Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$\hat{i}\cdot(\hat{j}\times\hat{k})+\hat{j}\cdot(\hat{i}\times\hat{k})+\hat{k}\cdot(\hat{i}\times\hat{j})$
(a) $0$
(b) $-1$
(c) $1$
(d) $3$
›Reveal solutionSolution

Use the standard unit-vector cross products $\hat{j}\times\hat{k}=\hat{i}$ , $\hat{i}\times\hat{k}=-\hat{j}$ , $\hat{i}\times\hat{j}=\hat{k}$ .

$\hat{i}\cdot(\hat{j}\times\hat{k})=\hat{i}\cdot\hat{i}=1$

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 mark
$[2\hat i,\ 3\hat i,\ \hat j] =$
›Reveal solutionSolution

A scalar triple product is zero whenever two of the three vectors are parallel (scalar multiples of each other).

$[2\hat i,\,3\hat i,\,\hat j]=2\hat i\cdot(3\hat i\times\hat j)$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$(2\vec{i} + 3\vec{k}) \cdot (\vec{i} + \vec{j} + 4\vec{k}) \times (3\vec{i} + \vec{j} + 7\vec{k}) =$
(a) $0$
(b) $112$
(c) $126$
(d) $192$
›Reveal solutionSolution

The scalar triple product evaluates to $0$ (the three vectors are coplanar).

First, $(\hat i+\hat j+4\hat k)\times(3\hat i+\hat j+7\hat k)$ :

$\hat i:(1)(7)-(4)(1)=3$ ; $\hat j:-[(1)(7)-(4)(3)]=5$ ; $\hat k:(1)(1)-(1)(3)=-2$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$(\vec{i} + \vec{j} + \vec{k}) \cdot (\vec{i} - \vec{j} - \vec{k}) \times (\vec{i} + 2\vec{j} - \vec{k}) =$
(a) $0$
(b) $2$
(c) $4$
(d) $6$
›Reveal solutionSolution

The scalar triple product equals $6$ .

First compute $(\hat i-\hat j-\hat k)\times(\hat i+2\hat j-\hat k)$ :

…

Start your 14-day free trial to see the full solution →
CBSE 2021Set I1 markMCQ
$\vec{j}\cdot(\vec{k}\times\vec{i}) =$
(a) $0$
(b) $1$
(c) $-1$
(d) $\vec{j}$
›Reveal solutionSolution

$\vec k\times\vec i=\vec j$ and $\vec j\cdot\vec j=1$ .

Using the right-handed cyclic rule, $\vec k\times\vec i=\vec j$ .

Then $\vec j\cdot(\vec k\times\vec i)=\vec j\cdot\vec j=1$ .

…

Start your 14-day free trial to see the full solution →

Q.The vectors 3i^−j^+2k^3\hat{i} - \hat{j} + 2\hat{k}3i^−j^​+2k^, 2i^+j^+3k^2\hat{i} + \hat{j} + 3\hat{k}2i^+j^​+3k^ and i^+λj^−k^\hat{i} + \lambda\hat{j} - \hat{k}i^+λj^​−k^ are coplanar if value of λ\lambdaλ is (A) −2-2−2 (B) 000 (C) 222 (D) Any real number

🎓Unlock everything free for 14 days

Q.The vectors $3\hat{i} - \hat{j} + 2\hat{k}$ , $2\hat{i} + \hat{j} + 3\hat{k}$ and $\hat{i} + \lambda\hat{j} - \hat{k}$ are coplanar if value of $\lambda$ is (A) $-2$ (B) $0$ (C) $2$ (D) Any real number

Unlock everything free for 14 days