For any vector x, where i, j, k have their usual meanings the value of (x × i)^2+(x × j)^2+(x × k)^2 where i, j, k have their usual meanings, is equal to

Correct answer: (B) 2|x|^2. Let x=α i+β j+γ k Then, x × i=-β k+γ j x × j =k-γ i x × k =-α j+β i Now, (x × i)^2 =(x × i) ·(x × i) =(-β k+γ j) ·(-β k+γ j) =β^2+γ^2 Similarly, (x × j)^2=α^2+γ^2 and (x × k)^2=α^2+β^2 ∴(x × i)^2+(x × j)^2+(x × k )^2 =β^2+γ^2+α^2+γ^2+α^2+β^2 =2 (α^2+β^2+γ^2 )=2|x|^2

Q102 / 155 · WBJEE 2017 · Mathematics

For any vector $\mathbf{x},$ where $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$ have their usual meanings the value of $(\mathbf{x} \times \hat{\mathbf{i}})^{2}+(\mathbf{x} \times \mathbf{j})^{2}+(\mathbf{x} \times \mathbf{k})^{2}$ where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ have their usual meanings, is equal to

A
$|\mathbf{x}|^{2}$
Correct answer
B
$2|\mathbf{x}|^{2}$
C
$3|\mathbf{x}|^{2}$
D
$4|\mathbf{x}|^{2}$

Solution

Let $\mathbf{x}=\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k}$ Then, $\mathbf{x} \times \mathbf{i}=-\beta \hat{\mathbf{k}}+\gamma \hat{\mathbf{j}}$ $\begin{aligned} \mathbf{x} \times \hat{\mathbf{j}} &=\hat{\mathbf{k}}-\gamma \hat{\mathbf{i}} \\ \mathbf{x} \times \mathbf{k} &=-\alpha \hat{\mathbf{j}}+\beta \hat{\mathbf{i}} \end{aligned}$ $\begin{aligned} \text {Now, }(\mathbf{x} \times \mathbf{i})^{2} &=(\mathbf{x} \times \mathbf{i}) \cdot(\mathbf{x} \times \hat{\mathbf{i}}) \\ &=(-\beta \hat{\mathbf{k}}+\gamma \hat{\mathbf{j}}) \cdot(-\beta \hat{\mathbf{k}}+\gamma \hat{\mathbf{j}}) \\ &=\beta^{2}+\gamma^{2} \end{aligned}$ Similarly, $(\mathbf{x} \times \mathbf{j})^{2}=\alpha^{2}+\gamma^{2}$ and $(\mathbf{x} \times \mathbf{k})^{2}=\alpha^{2}+\beta^{2}$ $\therefore(\mathbf{x} \times \mathbf{i})^{2}+(\mathbf{x} \times \mathbf{j})^{2}+(\mathbf{x} \times \mathbf{\mathbf { k }})^{2}$ $=\beta^{2}+\gamma^{2}+\alpha^{2}+\gamma^{2}+\alpha^{2}+\beta^{2}$ $=2\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)=2|\mathbf{x}|^{2}$

Hence, the correct answer is option: 'B'.

Solution

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