Choose the correct answer from the given four options. For any vector a , the value of ( a i )^2+( a j )^2+( a k )^2 is: 1. a ^2 2. 3 a ^2 3. 4 a ^2 4. 2 a ^2

4. 2 a ^2 Solution: Let a ^2=x i +y j +z k a ^2=x^2+y^2+z^2 a i i & j & k &y&z 1&0&0 = i [0]- j [-z]+ k [-y] =z j -y k ( a i )=(z j -y k )(z j -y k ) =y^2+z^2 Similarly, ( a j )^2=x^2+z^2 And ( a k )^2=x^2+y^2 ( a i )^2+( a j )^2+( a k )^2 =y^2+z^2+x^2+z^2+x^2+y^2 2(x^2+y^2+z^2)=2 a ^2

Choose the correct answer from the given four options. For any ve…

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Question

Choose the correct answer from the given four options.
For any vector $\vec{\text{a}},$ the value of $(\vec{\text{a}}\times\hat{\text{i}})^2+(\vec{\text{a}}\times\hat{\text{j}})^2+(\vec{\text{a}}\times\hat{\text{k}})^2$ is:

$\vec{\text{a}}^2$
$3\vec{\text{a}}^2$
$4\vec{\text{a}}^2$
$2\vec{\text{a}}^2$

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Answer

$2\vec{\text{a}}^2$

Solution:

Let $\vec{\text{a}}^2=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$

$\therefore\vec{\text{a}}^2=\text{x}^2+\text{y}^2+\text{z}^2$

$\therefore\ \vec{\text{a}}\times\vec{\text{i}}\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\\text{x}&\text{y}&\text{z}\\1&0&0 \end{vmatrix}$

$=\hat{\text{i}}[0]-\hat{\text{j}}[-\text{z}]+\hat{\text{k}}[-\text{y}]$

$=\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}}$

$\therefore\ (\vec{\text{a}}\times\hat{\text{i}})=(\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}})(\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}})$

$=\text{y}^2+\text{z}^2$

Similarly, $(\vec{\text{a}}\times\hat{\text{j}})^2=\text{x}^2+\text{z}^2$

And $(\vec{\text{a}}\times\hat{\text{k}})^2=\text{x}^2+\text{y}^2$

$\therefore\ (\vec{\text{a}}\times\hat{\text{i}})^2+(\vec{\text{a}}\times\hat{\text{j}})^2+(\vec{\text{a}}\times\hat{\text{k}})^2$ $=\text{y}^2+\text{z}^2+\text{x}^2+\text{z}^2+\text{x}^2+\text{y}^2$

$2(\text{x}^2+\text{y}^2+\text{z}^2)=2\vec{\text{a}}^2$

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