Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathbf{a}=\mathbf{i}+\mathbf{j}-\mathbf{k}, \mathbf{b}=\mathbf{i}-\mathbf{j}+\mathbf{k}, \mathbf{c}=\mathbf{i}-\mathbf{j}-\mathbf{k}$, then $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})$ is
Options:
Solution:
2805 Upvotes
Verified Answer
The correct answer is:
$2 \mathbf{i}-2 \mathbf{j}$
$\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(\mathbf{a} . \mathbf{c}) \mathbf{b}-(\mathbf{a} . \mathbf{b}) \mathbf{c}$
$\begin{aligned} & \ \mathbf{a} \cdot \mathbf{c}=(\mathbf{i}+\mathbf{j}-\mathbf{k}) \cdot(\mathbf{i}-\mathbf{j}-\mathbf{k})=1-1+1=1 \\ & \mathbf{a} \cdot \mathbf{b}=(\mathbf{i}+\mathbf{j}-\mathbf{k}) \cdot(\mathbf{i}-\mathbf{j}+\mathbf{k})=1-1-1=-1 \\ & \therefore \begin{array}{l}\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(1) \mathbf{b}-(-1) \mathbf{c} \\ \quad=\mathbf{b}+\mathbf{c}=\mathbf{i}-\mathbf{j}+\mathbf{k}+\mathbf{i}-\mathbf{j}-\mathbf{k}=2 \mathbf{i}-2 \mathbf{j}\end{array}\end{aligned}$
$\begin{aligned} & \ \mathbf{a} \cdot \mathbf{c}=(\mathbf{i}+\mathbf{j}-\mathbf{k}) \cdot(\mathbf{i}-\mathbf{j}-\mathbf{k})=1-1+1=1 \\ & \mathbf{a} \cdot \mathbf{b}=(\mathbf{i}+\mathbf{j}-\mathbf{k}) \cdot(\mathbf{i}-\mathbf{j}+\mathbf{k})=1-1-1=-1 \\ & \therefore \begin{array}{l}\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(1) \mathbf{b}-(-1) \mathbf{c} \\ \quad=\mathbf{b}+\mathbf{c}=\mathbf{i}-\mathbf{j}+\mathbf{k}+\mathbf{i}-\mathbf{j}-\mathbf{k}=2 \mathbf{i}-2 \mathbf{j}\end{array}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.