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Physics Question on Motion in a plane

Given two vectors A⃗=−i^+2j^−3k^\vec{ A }=-\hat{ i }+2 \hat{ j }-3 \hat{ k } and B⃗=4i^−2j^+6k^\vec{ B }=4 \hat{ i }-2 \hat{ j }+6 \hat{ k }. The angle made by (A⃗+B⃗)(\vec{ A }+\vec{ B }) with xx-axis is

A

30∘30^{\circ}

B

45∘45^{\circ}

C

60∘60^{\circ}

D

90∘90^{\circ}

Answer

45∘45^{\circ}

Explanation

Solution

Given, A⃗=−i^+2j^−3k^\vec{ A }=-\hat{ i }+2 \hat{ j }-3 \hat{ k } B⃗=4i^−2j^+6k^\vec{ B }=4 \hat{ i }-2 \hat{ j }+6 \hat{ k } Let angle made by (A⃗+B⃗)(\vec{ A }+\vec{ B }) with x−x- axis is θ\theta. cos⁡θ=(A⃗+B⃗)⋅i^∣A⃗+B⃗∣⋅∣i^∣\cos \theta=\frac{(\vec{ A }+\vec{ B }) \cdot \hat{ i }}{|\vec{ A }+\vec{ B }| \cdot|\hat{ i }|} =(3i^+3k^)⋅i9+9⋅1=\frac{(3 \hat{ i }+3 \hat{ k }) \cdot i}{\sqrt{9+9 \cdot 1}} =332=12=\frac{3}{3 \sqrt{2}}=\frac{1}{\sqrt{2}} ∴θ=45∘\therefore \theta=45^{\circ}