Question

Let \(\vec{a}=-\hat{i}-\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1\) and \(\vec{a} \times \vec{b}=\hat{i}-\hat{j}\). Then \(\vec{a}-6 \vec{b}\) is equal to

• A. $3(\hat{i}-\hat{j}-\hat{k})$
• B. $3(\hat{i}-\hat{j}+\hat{k})$
• C. $3(\hat{i}+\hat{j}-\hat{k})$
• D. $3(\hat{i}+\hat{j}+\hat{k})$

Answer 1

The Correct Option is D

Given the vector equations and properties, solving for b involves using the dot product and cross product information provided. Once \(b\) is calculated, \(a−6b\) is found by direct subtraction and scalar multiplication. The solution involves algebraic manipulation of vector components and verifying each option to match the calculated result.

Answer 2

The correct answer is (D) : $3(\hat{i}+\hat{j}+\hat{k})$
$\vec{a}\times \vec{b}=(\hat{i}-\hat{j})$
Taking cross product with $\vec{a}$
$\Rightarrow \vec{a}\times (\vec{a}\times \vec{b})=\vec{a}\times (\hat{i}-\hat{j})$
$\Rightarrow (\vec{a}\cdot \vec{b})\vec{a}-(\vec{a}\cdot \vec{a})\vec{b}=\hat{i}+\hat{j}+2\hat{k}$
$\Rightarrow \vec{a}-3\vec{b}=\hat{i}+\hat{j}+2\hat{k}$
$\Rightarrow 2\vec{a}-6\vec{b}=2\hat{i}+2\hat{j}+4\hat{k}$
$\Rightarrow \vec{a}-6\vec{b}=3\hat{i}+3\hat{j}+3\hat{k}$