Let $\vec{x}, \vec{y}$ and $\vec{z}$ be three vectors each of magnitude $\sqrt{2}$ and the angle between each pair of them is $\frac{\pi}{3}$. If $\vec{a}$ is a nonzero vector perpendicular to $\vec{x}$ and $\vec{y} \times \vec{z}$ and $\vec{b}$ is a nonzero vector perpendicular to $\vec{y}$ and $\vec{z} \times \vec{x}$, then

(A) $\vec{b}=(\vec{b} \cdot \vec{z})(\vec{z}-\vec{x})$; (B) $\vec{a}=(\vec{a} \cdot \vec{y})(\vec{y}-\vec{z})$; (C) $\vec{a} \cdot \vec{b}=-(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})$

(D) $\vec{a}=(\vec{a} \cdot \vec{y})(\vec{z}-\vec{y})$

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

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Let

x, y

and

z

be three vectors each of magnitude

2

and the angle between each pair of them is

\frac{π}{3}

. If

a

is a nonzero vector perpendicular to

x

and

y \times z

and

b

is a nonzero vector perpendicular to

y

and

z \times x

, then

(A) $b = (b \cdot z) (z - x)$ ✓ Correct
(B) $a = (a \cdot y) (y - z)$ ✓ Correct
(C) $a \cdot b = - (a \cdot y) (b \cdot z)$ ✓ Correct
(D) $a = (a \cdot y) (z - y)$

Answer: (A), (B), (C)

Solution

Given

∣ x ∣ = ∣ y ∣ = ∣ z ∣ = 2

and the angle between any pair is

\frac{π}{3}

, we have

x \cdot y = y \cdot z = z \cdot x = 1

. Since

a

is perpendicular to both

x

and

y \times z

, we have

a ∥ x \times (y \times z) = (x \cdot z) y - (x \cdot y) z = y - z

. So

a = λ (y - z)

. Taking dot product with

y

a \cdot y = λ (2 - 1) = λ

, so

a = (a \cdot y) (y - z)

, making (B) correct. Similarly,

b = (b \cdot z) (z - x)

, making (A) correct. For (C),

a \cdot b = \frac{1}{λ μ} (y - z) \cdot (z - x) = \frac{1}{λ μ} (1 - 2 - 1 + 1) = - \frac{1}{λ μ} = - (a \cdot y) (b \cdot z)

, so (C) is correct.

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

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