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JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Vector Algebra

If

a

and

b

are vectors in space given by

a = \frac{i ^ - 2 j ^}{5}

and

b = \frac{2 i ^ + j ^ + 3 k ^}{14}

, then the value of

(2 a + b) \cdot [(a \times b) \times (a - 2 b)]

is

Answer: 5

Solution

Use the vector triple product identity:

(u \times v) \times w = (u \cdot w) v - (v \cdot w) u

.

Let

u = a

,

v = b

,

w = a - 2 b

.

(a \times b) \times (a - 2 b) = (a \cdot (a - 2 b)) b - (b \cdot (a - 2 b)) a

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

Since

a

and

b

are unit vectors (check:

∣ a ∣^{2} = \frac{1 + 4}{5} = 1

,

∣ b ∣^{2} = \frac{4 + 1 + 9}{14} = 1

), and

a \cdot b = \frac{2 - 2}{70} = 0

.

So the expression becomes:

(1 - 0) b - (0 - 2) a = b + 2 a

.

Then

(2 a + b) \cdot (b + 2 a) = (2 a + b) \cdot (2 a + b) = ∣2 a + b ∣^{2} = 4∣ a ∣^{2} + 4 a \cdot b + ∣ b ∣^{2} = 4 + 0 + 1 = 5

.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer

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