Question 15 Marks
For any vector $\vec{a}$, prove that :
$|\overrightarrow{ a } \times \hat{ i }|^2+|\overrightarrow{ a } \times \hat{ j }|^2+|\overrightarrow{ a } \times \hat{ k }|^2= 2 |\overrightarrow{ a }|^2$
$|\overrightarrow{ a } \times \hat{ i }|^2+|\overrightarrow{ a } \times \hat{ j }|^2+|\overrightarrow{ a } \times \hat{ k }|^2= 2 |\overrightarrow{ a }|^2$
Answer
View full question & answer→Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$
${llrl} \therefore |\vec{a}| =\sqrt{a_1^2+a_2^2+a_3^2}$
$\text { or } |\vec{a}|^2 =a_1^2+a_2^2+a_3^2$
Therefore, $\vec{a} \times \hat{i}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \times \hat{i}$
$=a_1(\hat{i} \times \hat{i})+a_2(\hat{j} \times \hat{i})+a_3(\hat{k} \times \hat{i})$
$=a_1 \times 0+a_2(-\hat{k})+a_3 \hat{j}$
$=-a_2 \hat{k}+a_3 \hat{j}$
$\Rightarrow \quad|\vec{a} \times \hat{i}|=\sqrt{\left(-a_2\right)^2+\left(a_3\right)^2}=\sqrt{a_2^2+a_3^2}$
$\text { or } \quad|\vec{a} \times \hat{i}|^2=a_2^2+a_3^2 \ldots \ldots(1)$
$
\text { Similarly } \vec{a} \times \hat{j}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \times \hat{j}$
$=a_1 \hat{k}-a_3 \hat{i}$
${llrl} \therefore |\vec{a}| =\sqrt{a_1^2+a_2^2+a_3^2}$
$\text { or } |\vec{a}|^2 =a_1^2+a_2^2+a_3^2$
Therefore, $\vec{a} \times \hat{i}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \times \hat{i}$
$=a_1(\hat{i} \times \hat{i})+a_2(\hat{j} \times \hat{i})+a_3(\hat{k} \times \hat{i})$
$=a_1 \times 0+a_2(-\hat{k})+a_3 \hat{j}$
$=-a_2 \hat{k}+a_3 \hat{j}$
$\Rightarrow \quad|\vec{a} \times \hat{i}|=\sqrt{\left(-a_2\right)^2+\left(a_3\right)^2}=\sqrt{a_2^2+a_3^2}$
$\text { or } \quad|\vec{a} \times \hat{i}|^2=a_2^2+a_3^2 \ldots \ldots(1)$
$
\text { Similarly } \vec{a} \times \hat{j}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \times \hat{j}$
$=a_1 \hat{k}-a_3 \hat{i}$