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Physics Question on Motion in a plane

A unit vector in the direction of resultant vector of A⃗=−2i^+3j^+k^\vec{A}=-2{\hat{i}}+3\hat{j}+\hat{k} and B⃗=i^+2j^−4k^\vec{B}=\hat{i}+2\hat{j}-4\hat{k} is

A

−2i^+3j^+k^35\frac{-2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{35}}

B

i^+2j^−4k^35\frac{\hat{i}+2\hat{j}-4\hat{k}}{\sqrt{35}}

C

−i^+5j^−3k^35\frac{-\hat{i}+5\hat{j}-3\hat{k}}{\sqrt{35}}

D

−3i^+j^+5k^35\frac{-3\hat{i}+\hat{j}+5\hat{k}}{\sqrt{35}}

Answer

−i^+5j^−3k^35\frac{-\hat{i}+5\hat{j}-3\hat{k}}{\sqrt{35}}

Explanation

Solution

Here, A⃗=−2i^^+3j^+k^\vec{A}=-2\hat{\hat{i}}+3\hat{j}+\hat{k} B⃗=i^+2j^−4k^\vec{B}=\hat{i}+2\hat{j}-4\hat{k} The resultant vector of A⃗\vec{A} and B⃗\vec{B} is R⃗=A⃗+B⃗\vec{R}=\vec{A}+\vec{B} ∴R⃗=(−2i^+3j^+k^)+(i^+2j^−4k^)\therefore \vec{R}=\left(-2\hat{i}+3\hat{j}+\hat{k}\right)+\left(\hat{i}+2\hat{j}-4\hat{k}\right) =−i^+5j^−3k^=-\hat{i}+5\hat{j}-3\hat{k} ∣R⃗∣=(−1)2+(5)2+(−3)2\left|\vec{R}\right|=\sqrt{\left(-1\right)^{2}+\left(5\right)^{2}+\left(-3\right)^{2}} =1+25+9=\sqrt{1+25+9} =35=\sqrt{35} Unit vector in the direction of resultant vector of A⃗\vec{A} and B⃗\vec{B} is R^=R⃗∣R⃗∣\hat{R}=\frac{\vec{R}}{\left|\vec{R}\right|} =−i^+5j^−3k^35=\frac{-\hat{i}+5\hat{j}-3\hat{k}}{\sqrt{35}}