If a=hat{i}+2 hat{j}+3 hat{k},b=2 hat{i}+3 ha | vedclass

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(A) Any vector $a$ in $3D$ space can be expressed as a linear combination of three mutually orthogonal vectors $b$,$c$,and $b 	imes c$ (assuming $b$

Vedclass · Accepted Answer

$\sqrt{14}$

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(A) Any vector $a$ in $3D$ space can be expressed as a linear combination of three mutually orthogonal vectors $b$,$c$,and $b 	imes c$ (assuming $b$ and $c$ are non-parallel).The general expansion of a vector $a$ in terms of a basis ${b, c, b 	imes c}$ is given by:$a = \left\{\frac{a \cdot b}{|b|^2}ight\} b + \left\{\frac{a \cdot c}{|c|^2}ight\} c + \left\{\frac{a \cdot (b 	imes c)}{|b 	imes c|^2}ight\} (b 	imes c)$Comparing this with the given expression,we see that the expression is exactly equal to the vector $a$.Therefore,the magnitude of the expression is equal to the magnitude of vector $a$.$|a| = |\hat{i} + 2\hat{j} + 3\hat{k}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.

If $a=\hat{i}+2 \hat{j}+3 \hat{k}$,$b=2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $c$ is a vector perpendicular to $b$,then $\left\{\frac{a \cdot(b \times c)}{|b \times c|^2}\right\}(b \times c)+\left\{\frac{a \cdot b}{|b|^2}\right\} b+\left\{\frac{a \cdot c}{|c|^2}\right\} c$ is equal to:

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