Let a→ be a non-zero vector. If x→=i^×(a→×i^), y→=j^×(a→×j^)-a→ and z→=k^×(a→×k^)-a→, then x→ y→ z→ =

Correct answer: (C) 0. Given, x→=i^×(a→×i^), y→=j^×(a→×j^)-a→ and z→=k^×(a→×k^)-a→ ⇒x→=i^·i^a→-i^·a→i^, y→=j^·j^a→-j^·a→j^-a→ and z→=k^·k^a→-k^·a→k^-a→ ⇒x→=a→-i^·a→i^, y→=-j^·a→j^ and z→=-k^.a→k^ Let, a→=αi^+βj^+γk^, then x→=βj^+γk^, y→=-βj^ and z→=-γk^. Now, x→ y→ z→ =x→·y→×z→ ⇒ x→ y→ z→ =βj^+γk^·-βj^×-γk^ ⇒ x→ y→ z→ =βj^+γk^·βγi^=0

Q71 / 160 · TS EAMCET 2019 (03 May Shift 2) · Mathematics

Let $\vec{a}$ be a non-zero vector. If $\vec{x} = \hat{i} \times (\vec{a} \times \hat{i}), \vec{y} = \hat{j} \times (\vec{a} \times \hat{j}) - \vec{a}$ and $\vec{z} = \hat{k} \times (\vec{a} \times \hat{k}) - \vec{a},$ then $[\begin{array}{l} \vec{x} & \vec{y} & \vec{z} \end{array}] =$

A
$| \bar{a} |$
B
$2 | \bar{a} |$
Correct answer
C
$0$
D
$1$

Solution

Given, $\vec{x} = \hat{i} \times (\vec{a} \times \hat{i}), \vec{y} = \hat{j} \times (\vec{a} \times \hat{j}) - \vec{a}$ and $\vec{z} = \hat{k} \times (\vec{a} \times \hat{k}) - \vec{a}$

$\Rightarrow \vec{x} = (\hat{i} \cdot \hat{i}) \vec{a} - (\hat{i} \cdot \vec{a}) \hat{i}, \vec{y} = (\hat{j} \cdot \hat{j}) \vec{a} - (\hat{j} \cdot \vec{a}) \hat{j} - \vec{a}$ and $\vec{z} = (\hat{k} \cdot \hat{k}) \vec{a} - (\hat{k} \cdot \vec{a}) \hat{k} - \vec{a}$

$\Rightarrow \vec{x} = \vec{a} - (\hat{i} \cdot \vec{a}) \hat{i}, \vec{y} = - (\hat{j} \cdot \vec{a}) \hat{j}$ and $\vec{z} = - (\hat{k} . \vec{a}) \hat{k}$

Let, $\vec{a} = α \hat{i} + β \hat{j} + γ \hat{k}$ , then $\vec{x} = β \hat{j} + γ \hat{k}, \vec{y} = - β \hat{j}$ and $\vec{z} = - γ \hat{k}$ .

Now, $[\begin{array}{l} \vec{x} & \vec{y} & \vec{z} \end{array}] = \vec{x} \cdot (\vec{y} \times \vec{z})$

$\Rightarrow [\begin{array}{l} \vec{x} & \vec{y} & \vec{z} \end{array}] = (β \hat{j} + γ \hat{k}) \cdot ((- β) \hat{j} \times (- γ) \hat{k})$

$\Rightarrow [\begin{array}{l} \vec{x} & \vec{y} & \vec{z} \end{array}] = (β \hat{j} + γ \hat{k}) \cdot (β γ \hat{i}) = 0$

Hence, the correct answer is option: 'C'.

Let a→ be a non-zero vector. If x→=i^×(a→×i^), y→=j^×(a→×j^)-a→ and z→=k^×(a→×k^)-a→, then x→y→z→=

Solution

Relevant Questions

Let $\vec{a}$ be a non-zero vector. If $\vec{x} = \hat{i} \times (\vec{a} \times \hat{i}), \vec{y} = \hat{j} \times (\vec{a} \times \hat{j}) - \vec{a}$ and $\vec{z} = \hat{k} \times (\vec{a} \times \hat{k}) - \vec{a},$ then $[\begin{array}{l} \vec{x} & \vec{y} & \vec{z} \end{array}] =$