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CBSE Board Examination 2025 · Section A · 1 mark

Q11: Angle between b⃗\vec b and c⃗\vec c when a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0

Board 2025Ch 10 · Vector Algebra
The question

If a⃗+b⃗+c⃗=0⃗\vec a + \vec b + \vec c = \vec 0⁠, ∣a⃗∣=37|\vec a| = \sqrt{37}⁠, ∣b⃗∣=3|\vec b| = 3 and ∣c⃗∣=4|\vec c| = 4⁠, then angle between b⃗\vec b and c⃗\vec c is

(A) π6\frac{\pi}{6}

(B) π4\frac{\pi}{4}

(C) π3\frac{\pi}{3}

(D) π2\frac{\pi}{2}

Exam answer

b⃗+c⃗=−a⃗\vec b + \vec c = -\vec a⁠, so ∣b⃗+c⃗∣2=∣a⃗∣2|\vec b + \vec c|^2 = |\vec a|^2⁠.

Why?

We know lengths and want an angle, and the dot product links the two. Squaring a magnitude brings in a dot product, because ∣v⃗∣2=v⃗⋅v⃗|\vec v|^2 = \vec v \cdot \vec v⁠.

Also ∣−a⃗∣=∣a⃗∣\lvert -\vec a \rvert = |\vec a|⁠: reversing a vector does not change its length.

∣b⃗∣2+∣c⃗∣2+2 b⃗⋅c⃗=379+16+2 b⃗⋅c⃗=372 b⃗⋅c⃗=12b⃗⋅c⃗=6\begin{aligned} |\vec b|^2 + |\vec c|^2 + 2\,\vec b \cdot \vec c &= 37 \\ 9 + 16 + 2\,\vec b \cdot \vec c &= 37 \\ 2\,\vec b \cdot \vec c &= 12 \\ \vec b \cdot \vec c &= 6 \end{aligned}
Why?

Expand like a square of a sum:

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

The numbers: 32=93^2 = 9⁠, 42=164^2 = 16⁠, (37)2=37(\sqrt{37})^2 = 37⁠, and 37−25=1237 - 25 = 12⁠.

9+16=2525+2 b⃗⋅c⃗=372 b⃗⋅c⃗=37−25=12b⃗⋅c⃗=122=6\begin{aligned} 9 + 16 &= 25 \\ 25 + 2\,\vec b \cdot \vec c &= 37 \\ 2\,\vec b \cdot \vec c &= 37 - 25 = 12 \\ \vec b \cdot \vec c &= \frac{12}{2} = 6 \end{aligned}
cos⁡θ=b⃗⋅c⃗∣b⃗∣∣c⃗∣=63×4=12\cos\theta = \frac{\vec b \cdot \vec c}{|\vec b||\vec c|} = \frac{6}{3 \times 4} = \frac{1}{2}
Why?

This comes from b⃗⋅c⃗=∣b⃗∣∣c⃗∣cos⁡θ\vec b \cdot \vec c = |\vec b||\vec c|\cos\theta⁠, solved for cos⁡θ\cos\theta⁠. Put in the numbers:

cos⁡θ=6(3)(4)=612=6÷612÷6=12\begin{aligned} \cos\theta &= \frac{6}{(3)(4)} \\ &= \frac{6}{12} = \frac{6 \div 6}{12 \div 6} = \frac{1}{2} \end{aligned}

The angle between two vectors lies in [0,π][0, \pi]⁠, and there cos⁡θ=12\cos\theta = \frac12 only at θ=π3\theta = \frac{\pi}{3}⁠.

∴ θ=π3\therefore\ \theta = \frac{\pi}{3}

Answer: (C) π3\dfrac{\pi}{3}

Where this comes from

Chapter 10, Vector Algebra: Angle between vectors from a + b + c = 0.

Question type: Dot products with letters + The angle between two vectors (combines 2 types)

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: First, only lengths are given: write b⃗+c⃗=−a⃗\vec b + \vec c = -\vec a⁠, square both sides and expand, which gives 9+16+2 b⃗⋅c⃗=379 + 16 + 2\,\vec b\cdot\vec c = 37⁠, so b⃗⋅c⃗=6\vec b\cdot\vec c = 6⁠.

Type 4. The angle between two vectors

Spot it: "find the angle between the vectors".

Here: Then the angle: cos⁡θ=63×4=12\cos\theta = \frac{6}{3 \times 4} = \frac12⁠, so θ=π3\theta = \frac{\pi}{3}⁠.

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