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CBSE Sample Paper 2025-26 · Section A · 1 mark

Q20: r⃗×(a⃗+b⃗)=0⃗\vec r \times (\vec a + \vec b) = \vec 0 and r⃗\vec r perpendicular to a⃗×b⃗\vec a \times \vec b

Sample 2025-26Ch 10 · Vector Algebra
The question

Select the correct answer from the options (A), (B), (C) and (D) as given below.

(A) Both (A) and (R) are true and (R) is the correct explanation of (A).

(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).

(C) (A) is true but (R) is false.

(D) (A) is false but (R) is true.

Assertion(A): Given two non-zero vectors a⃗\vec a and b⃗\vec b⁠. If r⃗\vec r is another non-zero vector such that r⃗×(a⃗+b⃗)=0⃗\vec r \times (\vec a + \vec b) = \vec 0⁠. Then r⃗\vec r is perpendicular to a⃗×b⃗\vec a \times \vec b⁠.

Reason (R): The vector (a⃗+b⃗)(\vec a + \vec b) is perpendicular to the plane of a⃗\vec a and b⃗\vec b

Exam answer

r⃗×(a⃗+b⃗)=0⃗\vec r \times (\vec a + \vec b) = \vec 0⁠, so r⃗=λ(a⃗+b⃗)\vec r = \lambda(\vec a + \vec b)⁠, parallel to a⃗+b⃗\vec a + \vec b⁠.

Why?

For non-zero vectors, ∣p⃗×q⃗∣=∣p⃗∣ ∣q⃗∣sin⁡θ|\vec p \times \vec q| = |\vec p|\,|\vec q|\sin\theta⁠. This is 0 only when θ=0\theta = 0 or π\pi⁠, that is, when the vectors are parallel.

So the non-zero vector r⃗\vec r is a scalar multiple of a⃗+b⃗\vec a + \vec b⁠.

r⃗⋅(a⃗×b⃗)=λ[a⃗⋅(a⃗×b⃗)+b⃗⋅(a⃗×b⃗)]=λ(0+0)=0\begin{aligned} &\vec r \cdot (\vec a \times \vec b) \\ &= \lambda[\vec a \cdot (\vec a \times \vec b) + \vec b \cdot (\vec a \times \vec b)] \\ &= \lambda(0 + 0) = 0 \end{aligned}
Why?

The dot product distributes over addition, and λ\lambda comes outside.

a⃗×b⃗\vec a \times \vec b is perpendicular to both a⃗\vec a and b⃗\vec b⁠, so both dot products are 0:

a⃗⋅(a⃗×b⃗)=0b⃗⋅(a⃗×b⃗)=0\begin{aligned} \vec a \cdot (\vec a \times \vec b) &= 0 \\ \vec b \cdot (\vec a \times \vec b) &= 0 \end{aligned}

So r⃗\vec r is perpendicular to a⃗×b⃗\vec a \times \vec b⁠, and Assertion (A) is true.

Why?

For non-zero vectors, a zero dot product means the two vectors are perpendicular.

If a⃗+b⃗=0⃗\vec a + \vec b = \vec 0⁠, every r⃗\vec r satisfies the equation, but then b⃗=−a⃗\vec b = -\vec a and a⃗×b⃗=0⃗\vec a \times \vec b = \vec 0⁠, so r⃗⋅(a⃗×b⃗)=0\vec r \cdot (\vec a \times \vec b) = 0 still.

a⃗+b⃗\vec a + \vec b lies in the plane of a⃗\vec a and b⃗\vec b⁠, so it is not perpendicular to that plane. Reason (R) is false.

Why?

By the parallelogram law, a⃗+b⃗\vec a + \vec b is a diagonal of the parallelogram on a⃗\vec a and b⃗\vec b⁠, so it lies in their plane. The direction perpendicular to that plane is a⃗×b⃗\vec a \times \vec b⁠.

For example, with a⃗=i^\vec a = \hat i and b⃗=j^\vec b = \hat j⁠, a⃗+b⃗=i^+j^\vec a + \vec b = \hat i + \hat j lies in the xy-plane, while a⃗×b⃗=k^\vec a \times \vec b = \hat k⁠.

A is true and R is false, which is code (C).

Answer: (C) Assertion (A) is true but Reason (R) is false.

Where this comes from

Chapter 10, Vector Algebra: Vector product: parallel and perpendicular vectors.

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