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Mathematics Question on Vector Algebra

Let u→,v→,w→\overrightarrow{u},\overrightarrow{v},\overrightarrow{w} be such that ∣u→∣=1,∣v→∣=2,∣w→∣=3|\overrightarrow{u}|=1,|\overrightarrow{v}|=2,|\overrightarrow{w} |=3 . If the projection v→\overrightarrow{v} along u→\overrightarrow{u} is equal to that of w→\overrightarrow{w} along u→\overrightarrow{u} and v→\overrightarrow{v} and w→\overrightarrow{w} are ⊥\bot to each other , then ∣u→−v→+w→∣|\overrightarrow{u} -\overrightarrow{v} +\overrightarrow{w} | equals

A

14\sqrt{14}

B

7\sqrt{7}

C

2

D

14

Answer

14\sqrt{14}

Explanation

Solution

∣u⃗∣=1,∣v⃗∣=2,∣w⃗∣=3\left|\vec{u}\right|=1, \left|\vec{v}\right|=2, \left|\vec{w}\right|=3. By the given condition, ⇒v⃗ . u⃗∣u⃗∣=w⃗ . v⃗∣u⃗∣\Rightarrow \frac{\vec{v}\,.\,\vec{u}}{\left|\vec{u}\right|}=\frac{\vec{w}\,.\,\vec{v}}{\left|\vec{u}\right|} ⇒v⃗ . u⃗=w⃗ . u⃗\Rightarrow \vec{v}\,.\,\vec{u}=\vec{w}\,.\,\vec{u} Also, v⃗ . w⃗=0(∵v⃗ ⊥ w⃗)\vec{v}\,.\,\vec{w}=0\quad\left(\because \vec{v}\,\bot\,\vec{w}\right) Now ∣u⃗−v+w⃗∣2=u⃗2+v⃗2+w⃗2\left|\vec{u}-v+\vec{w}\right|^{2}=\vec{u}^{2}+\vec{v}^{2}+\vec{w}^{2} −2 u⃗ . v⃗−2 v⃗ . w⃗+2 w⃗ . u⃗-2\,\vec{u}\,.\,\vec{v}-2\,\vec{v}\,.\,\vec{w}+2\,\vec{w}\,.\,\vec{u} =1+4+9−2 v⃗⋅w⃗(∵v⃗⋅b⃗=u⃗⋅w⃗)=1+4+9-2\,\vec{v}\cdot\vec{w}\quad\left(\because \vec{v}\cdot\vec{b}=\vec{u}\cdot\vec{w}\right) =14−2(0)=14=14-2\left(0\right)=14 ∴∣u⃗−v⃗+w⃗∣=14\therefore \left|\vec{u}-\vec{v}+\vec{w}\right|=\sqrt{14}