MHT-CET Full Test-7 Mathematics Que-35 Solution

Q.35. \bar{u}, \bar{v},\bar{w} are three vectors such that | \bar{u}| =1, | \bar{v}| =2, | \bar{w}| =3. If the projection of \bar{v} along \bar{u} is equal to the projection of \bar{w} along \bar{u} and \bar{v} , \bar{w} are perpendicular to each other, then |\bar{u} - \bar{v} + \bar{w}| =

A. 4

B. \sqrt{7}

C. \sqrt{14}

D. 2

Answer :- C. \sqrt{14}

Explanation :-

|\vec{u}| = 1, |\vec{v}| = 2, |\vec{w}| = 3

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

\therefore\quad{\frac{\mathrm{v}\cdot{\overline{u}}}{|{\overline{u}}|}}={\frac{{\overline{w}}\cdot{\overline{u}}}{|{\overline{u}}|}} \therefore\quad\mathbf{\overline{v}}\cdot\mathbf{\overline{u}}=\mathbf{\overline{w}}\cdot\mathbf{\overline{u}} \therefore\quad(\overline{\mathrm{w}}-\overline{\mathrm{v}})\cdot\overline{\mathrm{u}}=0\quad….(\mathrm{i}) \mathrm{Now~consider,~|\overline{u}-\overline{v}+\overline{w}|=\sqrt{|\overline{u}+\overline{w}-\overline{v}|^{2}}} =\sqrt{|\overline{\mathrm{u}}|^{2}+|\overline{\mathrm{w}}-\overline{\mathrm{v}}|^{2}+2\overline{\mathrm{u}}\cdot(\overline{\mathrm{w}}-\overline{\mathrm{v}})}

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

=\sqrt{1+|\overline{\mathrm{w}}|^{2}+|\overline{\mathrm{v}}|^{2}-2(\overline{\mathrm{w}}\cdot\overline{\mathrm{v}})} =\sqrt{1+9+4+0}\quad\ldots[\because\overline{w}\mathrm{~and~}\overline{v}\text{ are perpendicular}] =\sqrt{14}