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(C) Given,$|\overrightarrow{u}|=1, |\overrightarrow{v}|=1, \overrightarrow{u} \cdot \overrightarrow{v} 
eq 0, \overrightarrow{u} \cdot \overrightarro

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Given,$|\overrightarrow{u}|=1, |\overrightarrow{v}|=1, \overrightarrow{u} \cdot \overrightarrow{v} 
eq 0, \overrightarrow{u} \cdot \overrightarrow{w}=1, \overrightarrow{v} \cdot \overrightarrow{w}=1, \overrightarrow{w} \cdot \overrightarrow{w}=4$.The volume of the parallelepiped is given by the scalar triple product $|[\overrightarrow{u} \overrightarrow{v} \overrightarrow{w}]| = \sqrt{2}$.Thus,$[\overrightarrow{u} \overrightarrow{v} \overrightarrow{w}]^2 = 2$.Using the property of the Gram determinant:$[\overrightarrow{u} \overrightarrow{v} \overrightarrow{w}]^2 = \begin{vmatrix} \overrightarrow{u} \cdot \overrightarrow{u} & \overrightarrow{u} \cdot \overrightarrow{v} & \overrightarrow{u} \cdot \overrightarrow{w} \ \overrightarrow{v} \cdot \overrightarrow{u} & \overrightarrow{v} \cdot \overrightarrow{v} & \overrightarrow{v} \cdot \overrightarrow{w} \ \overrightarrow{w} \cdot \overrightarrow{u} & \overrightarrow{w} \cdot \overrightarrow{v} & \overrightarrow{w} \cdot \overrightarrow{w} \end{vmatrix} = \begin{vmatrix} 1 & \overrightarrow{u} \cdot \overrightarrow{v} & 1 \ \overrightarrow{u} \cdot \overrightarrow{v} & 1 & 1 \ 1 & 1 & 4 \end{vmatrix} = 2$.Expanding the determinant:$1(4 - 1) - (\overrightarrow{u} \cdot \overrightarrow{v})(4\overrightarrow{u} \cdot \overrightarrow{v} - 1) + 1(\overrightarrow{u} \cdot \overrightarrow{v} - 1) = 2$.$3 - 4(\overrightarrow{u} \cdot \overrightarrow{v})^2 + \overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{u} \cdot \overrightarrow{v} - 1 = 2$.$-4(\overrightarrow{u} \cdot \overrightarrow{v})^2 + 2(\overrightarrow{u} \cdot \overrightarrow{v}) = 0$.$2(\overrightarrow{u} \cdot \overrightarrow{v})(1 - 2(\overrightarrow{u} \cdot \overrightarrow{v})) = 0$.Since $\overrightarrow{u} \cdot \overrightarrow{v} 
eq 0$,we have $\overrightarrow{u} \cdot \overrightarrow{v} = \frac{1}{2}$.Now,$|3\overrightarrow{u} + 5\overrightarrow{v}|^2 = 9|\overrightarrow{u}|^2 + 25|\overrightarrow{v}|^2 + 30(\overrightarrow{u} \cdot \overrightarrow{v}) = 9(1) + 25(1) + 30(\frac{1}{2}) = 9 + 25 + 15 = 49$.Therefore,$|3\overrightarrow{u} + 5\overrightarrow{v}| = \sqrt{49} = 7$.

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