CBSE Sample Paper 2022-23 · Section A · 1 mark

Q17: $|\vec a - 2\vec b|$ from $|\vec a|$ ⁠, $|\vec b|$ and $\vec a \cdot \vec b$

The question

(a) $\sqrt 2$

(b) $2\sqrt 6$

(d) $2\sqrt 2$

Exam answer

\begin{aligned} &|\vec a - 2\vec b|^2 \\ &= (\vec a - 2\vec b) \cdot (\vec a - 2\vec b) \\ &= |\vec a|^2 - 4\,\vec a \cdot \vec b + 4|\vec b|^2 \end{aligned}

Why?

The square of a length is the dot product of the vector with itself, $|\vec v|^2 = \vec v \cdot \vec v$ ⁠.

Expand like a product of brackets:

\begin{aligned} &\vec a \cdot \vec a - 2\,\vec a \cdot \vec b \\ &\quad - 2\,\vec b \cdot \vec a + 4\,\vec b \cdot \vec b \end{aligned}

Since $\vec b \cdot \vec a = \vec a \cdot \vec b$ ⁠, the two middle terms add to $-4\,\vec a \cdot \vec b$ ⁠.

\begin{aligned} &= 2^2 - 4(4) + 4(3^2) \\ &= 4 - 16 + 36 = 24 \end{aligned}

Why?

\begin{aligned} 2^2 &= 4 \\ 4(4) &= 16 \\ 4(3^2) &= 4(9) = 36 \\[4pt] 4 - 16 &= -12 \\ -12 + 36 &= 24 \end{aligned}

$\therefore\ |\vec a - 2\vec b| = \sqrt{24} = 2\sqrt 6$

Why?

Watch out: 24 is the square of the length, which is option (c). Take the square root, and simplify it with $24 = 4 \times 6$ ⁠:

\sqrt{24} = \sqrt 4\,\sqrt 6 = 2\sqrt 6

Answer: (b) $2\sqrt 6$

Where this comes from

Chapter 10, Vector Algebra: Magnitude using the dot product.

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 $\hat i, \hat j, \hat k$ at all, just $\vec a$ ⁠, $\vec b$ ⁠, their lengths and a dot product.

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Vector Algebra in past papers: question 40 of 48 · see them all

Q17: ∣a⃗−2b⃗∣|\vec a - 2\vec b|∣a−2b∣ from ∣a⃗∣|\vec a|∣a∣⁠, ∣b⃗∣|\vec b|∣b∣ and a⃗⋅b⃗\vec a \cdot \vec ba⋅b