Question

If a= i^ -3j^ +k^ and b= xi^+j^ -y k^ are parallel.find the values of x and y

Answer 1

GIVEN :–

$\\ {\huge{.}} \: \bf a = \hat{i} - 3 \hat{j} + \hat{k}$

$\\ {\huge{.}} \: \bf b = x\hat{i} + \hat{j} - y \hat{k}$

• a & b are parallel.

TO FIND :–

• Value of 'x' & 'y' = ?

SOLUTION :–

• We know that If two vectors a & b are parallel then –

$\\ \dashrightarrow \large \:{ \boxed{ \bf \vec{a} \times \vec{ b} =0 }}$

$\\ \implies \bf (\hat{i} - 3 \hat{j} + \hat{k}) \times (x\hat{i} + \hat{j} - y \hat{k}) = 0$

$\\ \implies\bf \left|\begin{array}{ccc} \bf \hat{i} & \bf \hat{j} & \bf\hat{k} \\ \\ \bf1 & \bf3 & \bf1 \\ \\ \bf x& \bf1 & \bf - y \end{array}\right| = 0$

$\\ \implies\bf \hat{i} \{(3)( - y) - (1)(1) \} - \hat{j} \{(1)( - y) - (1)(x) \} + \hat{k} \{ (1)(1) - (3)(x)\} = 0$

$\\ \implies\bf \hat{i} (- 3y - 1)- \hat{j} \{ - y-x \} + \hat{k} \{ 1 - 3x\} = 0$

$\\ \implies\bf \hat{i} (- 3y - 1) + \hat{j} \{ y+x \} + \hat{k} \{ 1 - 3x\} = 0$

• We should write this as –

$\\ \implies\bf \hat{i} (- 3y - 1) + \hat{j} \{ y+x \} + \hat{k} \{ 1 - 3x\} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}$

• Let's compare the coffieciant of $\:\: \bf \hat{i} \:\:$ –

$\\ \implies\bf - 3y - 1 = 0$

$\\ \implies\bf 3y + 1 = 0$

$\\ \implies \large{ \boxed{\bf y = \dfrac{ - 1}{3}}}$

• Now compare the coffieciant of $\:\: \bf \hat{k} \:\:$ –

$\\ \implies\bf 1 - 3x= 0$

$\\ \implies\bf 3x=1$

$\\ \implies \large{ \boxed{\bf x = \dfrac{1}{3}}}$

$\\ \rule{200}{4}$