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CBSE Sample Paper 2021-22 (Term 2) · Section B · 3 marks

Q9: If a⃗⋅b⃗=a⃗⋅c⃗\vec a\cdot\vec b = \vec a\cdot\vec c and a⃗×b⃗=a⃗×c⃗\vec a \times \vec b = \vec a \times \vec c with a⃗≠0⃗\vec a \ne \vec 0⁠, show b⃗=c⃗\vec b = \vec c

Sample 2021-22 T2Ch 10 · Vector Algebra
The question

If a⃗≠0⃗\vec a \ne \vec 0⁠, a⃗⋅b⃗=a⃗⋅c⃗\vec a\cdot\vec b = \vec a\cdot\vec c⁠, a⃗×b⃗=a⃗×c⃗\vec a \times \vec b = \vec a \times \vec c⁠, then show that b⃗=c⃗\vec b = \vec c⁠.

Exam answer

We have a⃗⋅b⃗=a⃗⋅c⃗\vec a\cdot\vec b = \vec a\cdot\vec c⁠.

⇒a⃗⋅(b⃗−c⃗)=0\Rightarrow \vec a\cdot(\vec b - \vec c) = 0
Why?

Move a⃗⋅c⃗\vec a\cdot\vec c to the left and use the distributive law: a⃗⋅b⃗−a⃗⋅c⃗=a⃗⋅(b⃗−c⃗)\vec a\cdot\vec b - \vec a\cdot\vec c = \vec a\cdot(\vec b - \vec c)⁠.

a⃗⋅b⃗=a⃗⋅c⃗a⃗⋅b⃗−a⃗⋅c⃗=0a⃗⋅(b⃗−c⃗)=0\begin{aligned} \vec a\cdot\vec b &= \vec a\cdot\vec c \\ \vec a\cdot\vec b - \vec a\cdot\vec c &= 0 \\ \vec a\cdot(\vec b - \vec c) &= 0 \end{aligned}

As a⃗≠0⃗\vec a \ne \vec 0⁠,

⇒b⃗−c⃗=0⃗ or a⃗⊥(b⃗−c⃗)⇒b⃗=c⃗ or a⃗⊥(b⃗−c⃗)\begin{aligned} &\Rightarrow \vec b - \vec c = \vec 0 \text{ or } \vec a \perp (\vec b - \vec c) \\ &\Rightarrow \vec b = \vec c \text{ or } \vec a \perp (\vec b - \vec c) \end{aligned}
Why?

For any vectors a⃗\vec a and d⃗\vec d at angle θ\theta⁠,

a⃗⋅d⃗=∣a⃗∣∣d⃗∣cos⁡θ\vec a\cdot\vec d = |\vec a||\vec d|\cos\theta

With d⃗=b⃗−c⃗\vec d = \vec b - \vec c and ∣a⃗∣≠0|\vec a| \ne 0⁠, this is zero only if d⃗=0⃗\vec d = \vec 0 or cos⁡θ=0\cos\theta = 0⁠, that is, a⃗⊥d⃗\vec a \perp \vec d⁠.

Also, a⃗×b⃗=a⃗×c⃗\vec a \times \vec b = \vec a \times \vec c⁠.

⇒a⃗×(b⃗−c⃗)=0⃗\Rightarrow \vec a \times (\vec b - \vec c) = \vec 0 ⇒b⃗−c⃗=0⃗ or a⃗∥(b⃗−c⃗)⇒b⃗=c⃗ or a⃗∥(b⃗−c⃗)\begin{aligned} &\Rightarrow \vec b - \vec c = \vec 0 \text{ or } \vec a \parallel (\vec b - \vec c) \\ &\Rightarrow \vec b = \vec c \text{ or } \vec a \parallel (\vec b - \vec c) \end{aligned}
Why?

In the same way, a⃗×b⃗−a⃗×c⃗=a⃗×(b⃗−c⃗)\vec a \times \vec b - \vec a \times \vec c = \vec a \times (\vec b - \vec c)⁠:

a⃗×b⃗=a⃗×c⃗a⃗×b⃗−a⃗×c⃗=0⃗a⃗×(b⃗−c⃗)=0⃗\begin{aligned} \vec a \times \vec b &= \vec a \times \vec c \\ \vec a \times \vec b - \vec a \times \vec c &= \vec 0 \\ \vec a \times (\vec b - \vec c) &= \vec 0 \end{aligned}

and

∣a⃗×d⃗∣=∣a⃗∣∣d⃗∣sin⁡θ|\vec a \times \vec d| = |\vec a||\vec d|\sin\theta

This is zero only if d⃗=0⃗\vec d = \vec 0 or sin⁡θ=0\sin\theta = 0⁠, that is, θ=0\theta = 0 or π\pi⁠: then a⃗∥d⃗\vec a \parallel \vec d⁠.

a⃗\vec a cannot be both perpendicular to (b⃗−c⃗)(\vec b - \vec c) and parallel to (b⃗−c⃗)(\vec b - \vec c)⁠.

Why?

Perpendicular needs θ=π2\theta = \frac{\pi}{2}⁠; parallel needs θ=0\theta = 0 or π\pi⁠. One angle cannot be both, so the only possibility left is b⃗−c⃗=0⃗\vec b - \vec c = \vec 0⁠.

Watch out: one condition alone is not enough. With a⃗=i^\vec a = \hat i⁠, b⃗=j^\vec b = \hat j⁠, c⃗=k^\vec c = \hat k⁠, both dot products are 0, but b⃗≠c⃗\vec b \ne \vec c⁠.

∴ b⃗=c⃗\therefore\ \vec b = \vec c⁠. Hence proved.

Where this comes from

Chapter 10, Vector Algebra: Dot and cross products: cancellation.

Quick lesson · Worked NCERT solutions · Interactive lesson

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