Question

X and Y components of a vector P are 3 units and 4 units respectively. If X and Y components of another vector (P+Q) are 5 and 8 units respectively then the angle made by Q with Y - axis is

Answer 1

Recquired angle = tan-1[1/2] = 26.56°

Answer 2

SOLUTION

GIVEN

X and Y components of a vector P are 3 units and 4 units respectively.

X and Y components of another vector (P+Q) are 5 and 8 units respectively

TO DETERMINE

The angle made by Q with Y - axis

EVALUATION

Here it is given that the X and Y components of a vector P are 3 units and 4 units respectively.

Thus we get

$\sf{\vec{P} = 3 \hat{i} + 4 \hat{j}}$

Again X and Y components of another vector (P+Q) are 5 and 8 units respectively

$\sf{\vec{P} + \vec{Q} = 5 \hat{i} + 8 \hat{j}}$

Thus we get

$\sf{\vec{Q} = (\vec{P} + \vec{Q} ) -\vec{P} }$

$\sf{ \implies \: \vec{Q} = ( 5 \hat{i} + 8 \hat{j}) - ( 3 \hat{i} + 4 \hat{j})}$

$\sf{ \implies \: \vec{Q} =2 \hat{i} + 4 \hat{j}}$

Let θ be the angle made by Q with Y - axis

Thus we get

$\displaystyle \sf{ \tan \theta = \frac{x- component \: \: of \: \vec{Q} }{y - component \:of \: \: \vec{Q} } }$

$\displaystyle \sf{ \implies \: \tan \theta = \frac{2}{4} }$

$\displaystyle \sf{ \implies \: \tan \theta = \frac{1}{2} }$

$\displaystyle \sf{ \implies \: \theta = {\tan}^{ - 1} \bigg( \frac{1}{2} \bigg) }$

FINAL ANSWER

Hence the required angle $\displaystyle \sf{ = {\tan}^{ - 1} \bigg( \frac{1}{2} \bigg) }$

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