MHT-CET Full Test-8 Mathematics Que-36 Solution

Q.36. \overrightarrow{u},\overrightarrow{v},\overrightarrow{w} are three vectors such that |\overrightarrow{u}| = 1, |\overrightarrow{v}| = 2, |\overrightarrow{w}| = 3. If the projection of \overrightarrow{v} along \overrightarrow{u} is equal to the projection of \overrightarrow{w} along \overrightarrow{u} and \overrightarrow{v}, \overrightarrow{w} are perpendicular to each other, then |\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}| = ?

A. 4

B. \sqrt{7}

C. \sqrt{14}

D. 2

Answer :- C. \sqrt{14}

Explanation:-

|\overline{\mathbf{u}}|=1, \left|\overline{\mathbf{v}}\right|=2, \left|\overline{\mathbf{w}}\right|=3

According to the given condition, (Projection of \overline{\mathbf{v}} along \overline{\mathbf{u}}) = (Projection of \overline{\mathbf{w}} along \overline{\mathbf{u}})

\frac{\overline{\mathbf{v}} \cdot \overline{\mathbf{u}}}{|\overline{\mathbf{u}}|} = \frac{\overline{\mathbf{w}} \cdot \overline{\mathbf{u}}}{|\overline{\mathbf{u}}|} \therefore \overline{\mathbf{v}} \cdot \overline{\mathbf{u}} = \overline{\mathbf{w}} \cdot \overline{\mathbf{u}}

\therefore (\overline{\mathbf{w}} - \overline{\mathbf{v}}) \cdot \overline{\mathbf{u}} = 0 ……(i)

Now consider |\overline{\mathbf{u}} - \overline{\mathbf{v}} + \overline{\mathbf{w}}| = \sqrt{(\overline{\mathbf{u}} + \overline{\mathbf{w}} - \overline{\mathbf{v}})^2}

= \sqrt{|\overline{\mathbf{u}}|^2 + |\overline{\mathbf{w}} - \overline{\mathbf{v}}|^2 + 2\overline{\mathbf{u}} \cdot (\overline{\mathbf{w}} - \overline{\mathbf{v}})}

= \sqrt{(1)^2 + |\overline{\mathbf{w}} - \overline{\mathbf{v}}|^2 + 0} ……[From (i)]

= \sqrt{1 + |\overline{\mathbf{w}}|^2 + |\overline{\mathbf{v}}|^2 - 2(\overline{\mathbf{w}} \cdot \overline{\mathbf{v}})}

= \sqrt{1 + 9 + 4 + 0} ……[: \overline{\mathbf{w}} and \overline{\mathbf{v}} are perpendicular]

= \sqrt{14}