Vector Algebra Question 1
Question 1 - 2024 (01 Feb Shift 1)
Let $\vec{a}=-5 \hat{i}+\hat{j}-3 \hat{k}, \vec{b}=\hat{i}+2 \hat{j}-4 \hat{k}$ and $\vec{c}=(((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i}$. Then $\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to
(1) -12
(2) -10
(3) -13
(4) -15
Show Answer
Answer (1)
Solution
$\vec{a}=-5 \hat{i}+j-3 \hat{k}$
$\vec{b}=\hat{i}+2 \hat{j}-4 \hat{k}$
$(\vec{a} \times \vec{b}) \times \hat{i}=(\vec{a} \cdot \hat{i}) \vec{b}-(\vec{b} \cdot \hat{i}) \vec{a}$
$=-5 \vec{b}-\vec{a}$
$=(((-5 \vec{b}-\vec{a}) \times \hat{i}) \times \hat{i})$
$=((-11 \hat{j}+23 \hat{k}) \times \hat{i}) \times \hat{i}$
$\Rightarrow(11 \hat{k}+23 \hat{j}) \times \hat{i}$
$\Rightarrow(11 \hat{j}-23 \hat{k})$
$\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})=11-23=-12$
