Let $\triangle P Q R$ be a triangle. Let $\vec{a}=\overrightarrow{Q R}, \vec{b}=\overrightarrow{R P}$ and $\vec{c}=\overrightarrow{P Q}$. If $|\vec{a}|=12,|\vec{b}|=4 \sqrt{3}$ and $\vec{b} \cdot \vec{c}=24$, then which of the following is (are) true?

(A) $\frac{|\vec{c}|^{2}}{2}-|\vec{a}|=12$; (C) $|\vec{a} \times \vec{b}+\vec{c} \times \vec{a}|=48 \sqrt{3}$; (D) $\vec{a} \cdot \vec{b}=-72$

(B) $\frac{|\vec{c}|^{2}}{2}+|\vec{a}|=30$

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Vector Algebra

Let

△ P QR

be a triangle. Let

a = QR, b = R P

and

c = P Q

. If

∣ a ∣ = 12, ∣ b ∣ = 43

and

b \cdot c = 24

, then which of the following is (are) true?

Answer: (A), (C), (D)

Note that

a + b + c = 0

(closed triangle).

Given:

∣ a ∣ = 12

∣ b ∣ = 43

b \cdot c = 24

.

From

c = - a - b

∣ c ∣^{2} = ∣ a + b ∣^{2} = ∣ a ∣^{2} + 2 a \cdot b + ∣ b ∣^{2} = 144 + 2 a \cdot b + 48

Also,

b \cdot c = b \cdot (- a - b) = - a \cdot b - ∣ b ∣^{2} = 24

- a \cdot b - 48 = 24

a \cdot b = - 72

✓ (D)

So:

∣ c ∣^{2} = 144 + 2 (- 72) + 48 = 144 - 144 + 48 = 48

(A)

\frac{∣ c ∣ ^{2}}{2} - ∣ a ∣ = \frac{48}{2} - 12 = 24 - 12 = 12

✓

(B)

\frac{∣ c ∣ ^{2}}{2} + ∣ a ∣ = 24 + 12 = 36 \neq = 30

✗

(C)

a \times b + c \times a = a \times b + c \times a = a \times (b - c) = a \times (b + a + b) = a \times (2 b + a) = 2 (a \times b)

∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ

. We have

a \cdot b = ∣ a ∣∣ b ∣ cos θ = - 72

12 \cdot 43 cos θ = - 72

cos θ = - \frac{72}{48 3} = - \frac{3}{2 3} = - \frac{3}{2}

θ = 150°, sin θ = \frac{1}{2}

∣ a \times b ∣ = 12 \cdot 43 \cdot \frac{1}{2} = 243

∣2 (a \times b) ∣ = 483

✓

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · One or more correct options