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CBSE Board Examination 2024 · Section B · 2 marks

Q24: If a⃗+b⃗⊥a⃗\vec a + \vec b \perp \vec a and 2a⃗+b⃗⊥b⃗2\vec a + \vec b \perp \vec b⁠, prove ∣b⃗∣=2 ∣a⃗∣|\vec b| = \sqrt2\,|\vec a|

Board 2024Ch 10 · Vector Algebra
The question

If a⃗\vec a and b⃗\vec b are two non-zero vectors such that (a⃗+b⃗)⊥a⃗(\vec a + \vec b) \perp \vec a and (2a⃗+b⃗)⊥b⃗(2\vec a + \vec b) \perp \vec b⁠, then prove that ∣b⃗∣=2 ∣a⃗∣|\vec b| = \sqrt2\,|\vec a|⁠.

Exam answer

(a⃗+b⃗)⊥a⃗(\vec a + \vec b) \perp \vec a⁠, so

(a⃗+b⃗)⋅a⃗=0⇒∣a⃗∣2+b⃗⋅a⃗=0…(i)\begin{aligned} (\vec a + \vec b)\cdot\vec a &= 0 \\ \Rightarrow |\vec a|^2 + \vec b\cdot\vec a &= 0 \quad\ldots(\text{i}) \end{aligned}
Why?

Two non-zero vectors are perpendicular exactly when their dot product is 0.

The dot product distributes over addition, and a vector dotted with itself is its length squared:

a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2 (a⃗+b⃗)⋅a⃗=a⃗⋅a⃗+b⃗⋅a⃗=∣a⃗∣2+b⃗⋅a⃗\begin{aligned} (\vec a + \vec b)\cdot\vec a &= \vec a\cdot\vec a + \vec b\cdot\vec a \\ &= |\vec a|^2 + \vec b\cdot\vec a \end{aligned}

(2a⃗+b⃗)⊥b⃗(2\vec a + \vec b) \perp \vec b⁠, so

(2a⃗+b⃗)⋅b⃗=0⇒2a⃗⋅b⃗+∣b⃗∣2=0…(ii)\begin{aligned} (2\vec a + \vec b)\cdot\vec b &= 0 \\ \Rightarrow 2\vec a\cdot\vec b + |\vec b|^2 &= 0 \quad\ldots(\text{ii}) \end{aligned}
Why?

The same two facts:

(2a⃗+b⃗)⋅b⃗=2 a⃗⋅b⃗+b⃗⋅b⃗\begin{aligned} &(2\vec a + \vec b)\cdot\vec b \\ &= 2\,\vec a\cdot\vec b + \vec b\cdot\vec b \end{aligned}

and b⃗⋅b⃗=∣b⃗∣2\vec b\cdot\vec b = |\vec b|^2⁠.

From (i), a⃗⋅b⃗=−∣a⃗∣2\vec a\cdot\vec b = -|\vec a|^2⁠. Using this in (ii),

Why?

The dot product is commutative, b⃗⋅a⃗=a⃗⋅b⃗\vec b\cdot\vec a = \vec a\cdot\vec b⁠, so (i) gives a⃗⋅b⃗\vec a\cdot\vec b in terms of ∣a⃗∣|\vec a|⁠. Putting it into (ii) removes the unknown a⃗⋅b⃗\vec a\cdot\vec b and leaves only the two lengths.

∣a⃗∣2+a⃗⋅b⃗=0a⃗⋅b⃗=−∣a⃗∣2(subtract ∣a⃗∣2)\begin{aligned} |\vec a|^2 + \vec a\cdot\vec b &= 0 \\ \vec a\cdot\vec b &= -|\vec a|^2 \quad (\text{subtract } |\vec a|^2) \end{aligned}
2(−∣a⃗∣2)+∣b⃗∣2=0⇒∣b⃗∣2=2∣a⃗∣2\begin{aligned} 2(-|\vec a|^2) + |\vec b|^2 &= 0 \\ \Rightarrow |\vec b|^2 &= 2|\vec a|^2 \end{aligned}
Why?

Open the bracket, then move the ∣a⃗∣2|\vec a|^2 term across:

2(−∣a⃗∣2)+∣b⃗∣2=0−2∣a⃗∣2+∣b⃗∣2=0∣b⃗∣2=2∣a⃗∣2(add 2∣a⃗∣2)\begin{aligned} 2(-|\vec a|^2) + |\vec b|^2 &= 0 \\ -2|\vec a|^2 + |\vec b|^2 &= 0 \\ |\vec b|^2 &= 2|\vec a|^2 \quad (\text{add } 2|\vec a|^2) \end{aligned}

A length is never negative, so we take the positive square root of both sides: 2∣a⃗∣2=2 ∣a⃗∣\sqrt{2|\vec a|^2} = \sqrt2\,|\vec a|⁠.

∣b⃗∣=2∣a⃗∣2=2×∣a⃗∣2=2 ∣a⃗∣\begin{aligned} |\vec b| &= \sqrt{2|\vec a|^2} \\ &= \sqrt2 \times \sqrt{|\vec a|^2} \\ &= \sqrt2\,|\vec a| \end{aligned}

∴ ∣b⃗∣=2 ∣a⃗∣\therefore\ |\vec b| = \sqrt2\,|\vec a|⁠. Hence proved.

Where this comes from

Chapter 10, Vector Algebra: Dot product: perpendicular vectors.

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: Each perpendicular pair gives a dot product of 0. Expand (a⃗+b⃗)⋅a⃗=0(\vec a + \vec b)\cdot\vec a = 0 and (2a⃗+b⃗)⋅b⃗=0(2\vec a + \vec b)\cdot\vec b = 0⁠, then put a⃗⋅b⃗=−∣a⃗∣2\vec a\cdot\vec b = -|\vec a|^2 from the first into the second.

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