Let \(\vec{a}, \vec{b}\), and \(\vec{c}\) be three non-coplanar unit vectors such that the angle between every pair of them is \(\frac{\pi}{3}\) . If \(\vec{a} \times \vec{b}+\vec{b} \times \vec{c}=p \vec{a}+q \vec{b}+r \vec{c}\), where \(p, q\) and \(r\) are scalars, then the value of \(\frac{p^{2}+2 q^{2}+r^{2}}{q^{2}}\) is

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Vector Algebra

Let

a, b

, and

c

be three non-coplanar unit vectors such that the angle between every pair of them is

\frac{π}{3}

. If

a \times b + b \times c = p a + q b + r c

, where

p, q

and

r

are scalars, then the value of

\frac{p ^{2} + 2 q ^{2} + r ^{2}}{q ^{2}}

Answer: 4

Given

∣ a ∣ = ∣ b ∣ = ∣ c ∣ = 1

and

a \cdot b = b \cdot c = c \cdot a = cos (\frac{π}{3}) = \frac{1}{2}

. Taking dot product of

a \times b + b \times c = p a + q b + r c

with

b

0 = p (\frac{1}{2}) + q (1) + r (\frac{1}{2})

, giving

p + 2 q + r = 0

. With

a

[a b c] = p (1) + q (\frac{1}{2}) + r (\frac{1}{2})

, giving

2 [a b c] = 2 p + q + r

. With

c

[a b c] = p (\frac{1}{2}) + q (\frac{1}{2}) + r (1)

, giving

2 [a b c] = p + q + 2 r

. Solving these three equations yields

p = - q = r

. Therefore

\frac{p ^{2} + 2 q ^{2} + r ^{2}}{q ^{2}} = \frac{q ^{2} + 2 q ^{2} + q ^{2}}{q ^{2}} = 4

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer