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CBSE Sample Paper 2022-23 · Section B · 2 marks

Q25: Find ∣x⃗∣|\vec x| if (x⃗−a⃗)⋅(x⃗+a⃗)=12(\vec x - \vec a)\cdot(\vec x + \vec a) = 12⁠, a⃗\vec a a unit vector

Sample 2022-23Ch 10 · Vector Algebra
The question

Find ∣x⃗∣|\vec x| if (x⃗−a⃗)⋅(x⃗+a⃗)=12(\vec x - \vec a)\cdot(\vec x + \vec a) = 12⁠, where a⃗\vec a is a unit vector.

Exam answer

Since a⃗\vec a is a unit vector, ∣a⃗∣=1|\vec a| = 1⁠.

(x⃗−a⃗)⋅(x⃗+a⃗)=12⇒ x⃗⋅x⃗+x⃗⋅a⃗−a⃗⋅x⃗−a⃗⋅a⃗=12\begin{aligned} &(\vec x - \vec a)\cdot(\vec x + \vec a) = 12 \\ \Rightarrow\ &\vec x\cdot\vec x + \vec x\cdot\vec a - \vec a\cdot\vec x \\ &\quad - \vec a\cdot\vec a = 12 \end{aligned}
Why?

A unit vector is a vector of magnitude 1.

The dot product is distributive, so the brackets multiply out like (p−q)(p+q)(p - q)(p + q) in algebra: each term of the first bracket is dotted with each term of the second.

As x⃗⋅a⃗=a⃗⋅x⃗\vec x\cdot\vec a = \vec a\cdot\vec x⁠,

Why?

The dot product is commutative, so the two middle terms are equal and cancel. Also, the dot product of a vector with itself is the square of its magnitude:

x⃗⋅x⃗=∣x⃗∣2\vec x\cdot\vec x = |\vec x|^2
∣x⃗∣2−∣a⃗∣2=12⇒ ∣x⃗∣2−1=12⇒ ∣x⃗∣2=13\begin{aligned} &|\vec x|^2 - |\vec a|^2 = 12 \\ \Rightarrow\ &|\vec x|^2 - 1 = 12 \\ \Rightarrow\ &|\vec x|^2 = 13 \end{aligned}
Why?

Put ∣a⃗∣2=12=1|\vec a|^2 = 1^2 = 1 and add 1 to both sides.

∣x⃗∣2−1=12∣x⃗∣2=12+1=13∣x⃗∣=13\begin{aligned} |\vec x|^2 - 1 &= 12 \\ |\vec x|^2 &= 12 + 1 = 13 \\ |\vec x| &= \sqrt{13} \end{aligned}

A magnitude is never negative, so take only the positive square root of 13.

∴ ∣x⃗∣=13\therefore\ |\vec x| = \sqrt{13}

Where this comes from

Chapter 10, Vector Algebra: Dot product: magnitude from (x − a)·(x + a).

Question type: Dot products with letters

How to spot it: click to see

Type 7. Dot products with letters: expand like brackets

Spot it: no i^,j^,k^\hat i, \hat j, \hat k at all, just a⃗\vec a⁠, b⃗\vec b⁠, their lengths and a dot product.

Here: No components, only a unit vector a⃗\vec a⁠: expand (x⃗−a⃗)⋅(x⃗+a⃗)(\vec x - \vec a)\cdot(\vec x + \vec a)⁠, the middle terms cancel, and ∣x⃗∣2−1=12|\vec x|^2 - 1 = 12⁠.

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