Question

A vector perpendicular to the vector (i+2j) and having magnitude 3root5

Answer 1

GIVEN :–

• A vector perpendicular to the vector (i+2j).

• Magnitude of vector is 3√5 .

TO FIND :–

• Vector = ?

SOLUTION :–

• Let a vector → $\: \: \bf x \: \hat{i} \: + \: y \: \hat{j}$

▪︎According to the first condition –

• We know that if two vectors $\bf \overrightarrow{P}$ and $\bf \overrightarrow{Q}$ are perpendicular then –

$\\ \large \implies{ \boxed{\bf\overrightarrow{P}. \overrightarrow{Q}= 0}}$

• So that –

$\\\implies \bf (x \: \hat{i} \: + \: y \: \hat{j}).(\: \hat{i} \: + \: 2\:\hat{j}) = 0$

$\\\implies \bf x + 2y= 0 \: \: \: \: - - - - eq.(1)$

• According to the second condition –

$\\\implies \bf Magnitude \: \: of \: \: vector = 3 \sqrt{5}$

$\\\implies \bf |x \: \hat{i} \: + \: y \: \hat{j}| = 3 \sqrt{5}$

$\\\implies \bf \sqrt{ {x}^{2} + {y}^{2} } = 3 \sqrt{5}$

$\\\implies \bf (\sqrt{ {x}^{2} + {y}^{2}})^{2} = (3 \sqrt{5})^{2}$

$\\\implies \bf {x}^{2} + {y}^{2}=9 \times 5$

$\\\implies \bf {x}^{2} + {y}^{2}=45$

• Using eq.(1) –

$\\\implies \bf {( - 2y)}^{2} + {y}^{2}=45$

$\\\implies \bf 4{y}^{2} + {y}^{2}=45$

$\\\implies \bf 5{y}^{2}=45$

$\\\implies \bf {y}^{2}=9$

$\\\implies \large{ \boxed{ \bf y= \pm3}}$

• And –

$\\\implies \large{ \boxed{ \bf x= \mp6}}$

• Hence , The vector equation –

$\\ \implies \bf\: \:Vector =6 \: \hat{i} \: - \: 3\: \hat{j} \: \: \: \: \: \: or \: \: \: \: - 6 \: \hat{i} \: + \: 3\: \hat{j}$