CBSE Sample Paper 2024-25 · Section A · 1 mark

Q10: $|\vec a - \vec b|$ when $|\vec a| = 3$ ⁠, $|\vec b| = 4$ ⁠, $|\vec a + \vec b| = 5$

Sample 2024-25Ch 10 · Vector Algebra

The question

If $|\vec a| = 3$ ⁠, $|\vec b| = 4$ and $|\vec a + \vec b| = 5$ ⁠, then $|\vec a - \vec b| =$

(A) $3$

(B) $4$

(D) $8$

Exam answer

Here $|\vec a + \vec b|^2 = 25$ ⁠, $|\vec a|^2 = 9$ and $|\vec b|^2 = 16$ ⁠.

We have

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

Why?

This comes from expanding each square with $|\vec v|^2 = \vec v \cdot \vec v$ ⁠:

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

Adding the two, the $\vec a \cdot \vec b$ terms cancel.

\begin{aligned} 25 + |\vec a - \vec b|^2 &= 2(9 + 16) = 50 \\ |\vec a - \vec b|^2 &= 25 \end{aligned}

Why?

Put in the three squares from the first line, then subtract 25 from both sides: $50 - 25 = 25$ ⁠.

\begin{aligned} 2(9 + 16) &= 2(25) = 50 \\ 25 + |\vec a - \vec b|^2 &= 50 \\ |\vec a - \vec b|^2 &= 50 - 25 = 25 \end{aligned}

$\therefore\ |\vec a - \vec b| = 5$

Why?

A magnitude is never negative, so take the positive square root of 25.

A quick picture: the first expansion gives $25 = 9 + 16 + 2\,\vec a \cdot \vec b$ ⁠, so $\vec a \cdot \vec b = 0$ ⁠. The vectors are perpendicular, and both diagonals of the 3 by 4 rectangle are 5 long.

Answer: (C) $5$

Where this comes from

Chapter 10, Vector Algebra: Magnitudes: |a + b| and |a − b|.

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.

Quick lesson · Worked NCERT solutions · Interactive lesson

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Vector Algebra in past papers: question 30 of 48 · see them all

Q10: ∣a⃗−b⃗∣|\vec a - \vec b|∣a−b∣ when ∣a⃗∣=3|\vec a| = 3∣a∣=3⁠, ∣b⃗∣=4|\vec b| = 4∣b∣=4⁠, ∣a⃗+b⃗∣=5|\vec a + \vec b| = 5∣a+b∣=5

Q10: $|\vec a - \vec b|$ when $|\vec a| = 3$ ⁠, $|\vec b| = 4$ ⁠, $|\vec a + \vec b| = 5$