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If $\mathbf{a}=2 \hat{\mathbf{j}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{c}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ are three vectors, then $|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=$
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2639 Upvotes
Verified Answer
The correct answer is:
$\sqrt{\frac{11}{39}}|\mathbf{a} \times(\mathbf{b} \times \mathbf{c})|$
We have,
$\begin{gathered}
\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \\
\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
\mathbf{c}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=|(\mathbf{c} \cdot \mathbf{a}) \mathbf{b}-(\mathbf{c} \cdot \mathbf{b}) \mathbf{a}| \\
=|(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(6 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})| \\
=|-4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+\hat{\mathbf{k}}|=\sqrt{16+49+1}=\sqrt{66} \\
\text { and }|\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=|(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \mid
\end{gathered}$
$\begin{array}{ll}
= & |(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(14 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})| \\
= & |-12 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}|=\sqrt{144+81+9}=\sqrt{234} \\
\Rightarrow & \quad \frac{|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|}{\mid \mathbf{a} \times(\mathbf{b} \times \mathbf{c})}=\sqrt{\frac{66}{234}} \\
\Rightarrow & \quad|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=\sqrt{\frac{11}{39}}|\mathbf{a} \times(\mathbf{b} \times \mathbf{c})|
\end{array}$
$\begin{gathered}
\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \\
\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
\mathbf{c}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=|(\mathbf{c} \cdot \mathbf{a}) \mathbf{b}-(\mathbf{c} \cdot \mathbf{b}) \mathbf{a}| \\
=|(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(6 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})| \\
=|-4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+\hat{\mathbf{k}}|=\sqrt{16+49+1}=\sqrt{66} \\
\text { and }|\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=|(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \mid
\end{gathered}$
$\begin{array}{ll}
= & |(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(14 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})| \\
= & |-12 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}|=\sqrt{144+81+9}=\sqrt{234} \\
\Rightarrow & \quad \frac{|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|}{\mid \mathbf{a} \times(\mathbf{b} \times \mathbf{c})}=\sqrt{\frac{66}{234}} \\
\Rightarrow & \quad|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=\sqrt{\frac{11}{39}}|\mathbf{a} \times(\mathbf{b} \times \mathbf{c})|
\end{array}$
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