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Let $\overrightarrow{a}=\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k}, \overrightarrow{b}= b_{1}\widehat{i}+b_{2}\widehat{j}+\sqrt{2}\widehat{k}$ and $\overrightarrow{c}=5\widehat{i}+\widehat{j}+\sqrt{2}\widehat{k},$ be three vectors such that the projection vector of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\overrightarrow{a}$ . If $\overrightarrow{a}+\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$ , then $|\overrightarrow{b}|$ is equal to :
Option 1)

6
Option 2)

4
Option 3)

$\sqrt{22}$
Option 4)

$\sqrt{32}$

Answers (1)

best_answer

Projection of vector a on vector b -

$\vec{a}\cos \Theta = \frac{\vec{a}.\vec{b}}{\left | \vec{b} \right |}$

- wherein

Projection of vector a on vector b

Angle between vector a and vector b -

$\cos \Theta =\frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}$

- wherein

Here $0\leq \Theta \leq \pi$ ??????

From the concept we have learnt

Projection of $\vec{b}$ on $\vec{a} = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|} = |\vec{a}|$

$\Rightarrow b_1 + b_2 = 1 \;\;\; - (1)$

and, $(\vec{a} + \vec{b}) \perp \vec{c} \Rightarrow (\vec{a} + \vec{b}) \cdot \vec{c} = 0$

$\Rightarrow 5b_1 + b_2= -10\;\;\;- (2)$

From (1) and (2)

$b_1 = -3$ and $b_2 = 5$

$|\vec{b}| = \sqrt{9 + 25 + 2} = 6$

Option 1)

6
Option 2)

4

Option 3)

$\sqrt{22}$

Option 4)

$\sqrt{32}$

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