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Let $\overrightarrow{\mathbf{a}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$
Assertion (A) : The identity
$|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}|^2+|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}|^2+|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}|^2=2|\overrightarrow{\mathbf{a}}|^2$ holds for
$\vec{a}$.
Reason (R) : $\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}=a_3 \hat{\mathbf{j}}-a_2 \hat{\mathbf{k}}$,
$\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}=a_1 \hat{\mathbf{k}}-a_3 \hat{\mathbf{i}}, \overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}=a_2 \hat{\mathbf{i}}-a_1 \hat{\mathbf{j}}$
Which of the following is correct?
Options:
Assertion (A) : The identity
$|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}|^2+|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}|^2+|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}|^2=2|\overrightarrow{\mathbf{a}}|^2$ holds for
$\vec{a}$.
Reason (R) : $\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}=a_3 \hat{\mathbf{j}}-a_2 \hat{\mathbf{k}}$,
$\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}=a_1 \hat{\mathbf{k}}-a_3 \hat{\mathbf{i}}, \overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}=a_2 \hat{\mathbf{i}}-a_1 \hat{\mathbf{j}}$
Which of the following is correct?
Solution:
1077 Upvotes
Verified Answer
The correct answer is:
Both $(\mathrm{A})$ and $(\mathrm{R})$ are true and $(\mathrm{R})$ is the correct reason for (A)
Given,
$\overrightarrow{\mathbf{a}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$
Now, $|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}|^2=(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}) \cdot(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}})$
$\begin{aligned} & =\left(a_3 \hat{\mathbf{j}}-a_2 \hat{\mathbf{k}}\right) \cdot\left(a_3 \hat{\mathbf{j}}-a_2 \hat{\mathbf{k}}\right) \\ & =a_3^2+a_2^2\end{aligned}$
$\begin{aligned}|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}|^2 & =\left(a_1 \hat{\mathbf{k}}-a_3 \hat{\mathbf{i}}\right) \cdot\left(a_1 \hat{\mathbf{k}}-a_3 \hat{\mathbf{i}}\right) \\ & =a_1^2+a_3^2\end{aligned}$
$\begin{aligned}|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}|^2 & =\left(a_1 \hat{\mathbf{i}}+a_1 \hat{\mathbf{j}}\right) \cdot\left(a_2 \hat{\mathbf{i}}-a_1 \hat{\mathbf{j}}\right) \\ & =a_2{ }^2+a_1{ }^2\end{aligned}$
Now,
$\begin{aligned}|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}|^2 & +|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}|^2+|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}|^2 \\ & =2\left(a_1^2+a_2^2+a_3^2\right) \\ & =2|\overrightarrow{\mathbf{a}}|^2\end{aligned}$
Hence, both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is correct reason for $\mathrm{A}$.
$\overrightarrow{\mathbf{a}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$
Now, $|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}|^2=(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}) \cdot(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}})$
$\begin{aligned} & =\left(a_3 \hat{\mathbf{j}}-a_2 \hat{\mathbf{k}}\right) \cdot\left(a_3 \hat{\mathbf{j}}-a_2 \hat{\mathbf{k}}\right) \\ & =a_3^2+a_2^2\end{aligned}$
$\begin{aligned}|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}|^2 & =\left(a_1 \hat{\mathbf{k}}-a_3 \hat{\mathbf{i}}\right) \cdot\left(a_1 \hat{\mathbf{k}}-a_3 \hat{\mathbf{i}}\right) \\ & =a_1^2+a_3^2\end{aligned}$
$\begin{aligned}|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}|^2 & =\left(a_1 \hat{\mathbf{i}}+a_1 \hat{\mathbf{j}}\right) \cdot\left(a_2 \hat{\mathbf{i}}-a_1 \hat{\mathbf{j}}\right) \\ & =a_2{ }^2+a_1{ }^2\end{aligned}$
Now,
$\begin{aligned}|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}|^2 & +|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}|^2+|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}|^2 \\ & =2\left(a_1^2+a_2^2+a_3^2\right) \\ & =2|\overrightarrow{\mathbf{a}}|^2\end{aligned}$
Hence, both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is correct reason for $\mathrm{A}$.
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