Let a→=i^-2j^+k^ and b→=i^-j^+k^, be two vectors. If c→, is a vector such that b→×c→=b→×a→ and c→⋅a→=0, then c→⋅b→, is equal to.

Correct answer: (C) -1/2. a→×b→×c→=a→×b→×a→ ⇒-a→.b→c→=a→.a→b→-a→.b→a→ Using given information a→=i^-2j^+k^ and b→=i^-j^+k^ , we can write a→.b→=4, a→.a→=6 ⇒-4c→=6i^-j^+k^-4i^-2j^+k^ ⇒-4c→=2i^+2j^+2k^ ⇒c→=-1/2i^+j^+k^ ⇒b→.c→=-1/2.

Q27 / 75 · JEE Main 2020 (08 Jan Shift 2) · Mathematics

Let $\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ , be two vectors. If $\vec{c}$ , is a vector such that $\vec{b} \times \vec{c} = \vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a} = 0,$ then $\vec{c} \cdot \vec{b}$ , is equal to.

A
$- \frac{3}{2}$
B
$\frac{1}{2}$
Correct answer
C
$- \frac{1}{2}$
D
$- 1$

Solution

$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{a} \times (\vec{b} \times \vec{a})$
$\Rightarrow - (\vec{a} . \vec{b}) \vec{c} = (\vec{a} . \vec{a}) \vec{b} - (\vec{a} . \vec{b}) \vec{a}$

Using given information $\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ , we can write $\vec{a} . \vec{b} = 4$ , $\vec{a} . \vec{a} = 6$
$\Rightarrow - 4 \vec{c} = 6 (\hat{i} - \hat{j} + \hat{k}) - 4 (\hat{i} - 2 \hat{j} + \hat{k})$
$\Rightarrow - 4 \vec{c} = 2 \hat{i} + 2 \hat{j} + 2 \hat{k}$
$\Rightarrow \vec{c} = - \frac{1}{2} (\hat{i} + \hat{j} + \hat{k})$
$\Rightarrow \vec{b} . \vec{c} = - \frac{1}{2}$ .

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

Let a→=i^-2j^+k^ and b→=i^-j^+k^, be two vectors. If c→, is a vector such that b→×c→=b→×a→ and c→⋅a→=0, then c→⋅b→, is equal to.

Solution

Relevant Questions

Let $\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ , be two vectors. If $\vec{c}$ , is a vector such that $\vec{b} \times \vec{c} = \vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a} = 0,$ then $\vec{c} \cdot \vec{b}$ , is equal to.