If A¯ = 4 i^ - 2j^ + 6 k^, and B¯ = i^ - 2j^ - 3 k^ , the angle which the A¯ +B¯ makes with the x – axis is

A→+B→=5i^−4j^+3k^(A→+B→).i^=A→+B→.i^.Cosθ⇒5=52+(−4)2+32.1.Cosθ⇒Cosθ=550,θ=Cos−1550

If A¯ = 4 i^ - 2j^ + 6 k^, and B¯ = i^ - 2j^ - 3 k^ , the angle which the A¯ +B¯ makes with the x – axis is

A→+B→=5i^−4j^+3k^(A→+B→).i^=A→+B→.i^.Cosθ⇒5=52+(−4)2+32.1.Cosθ⇒Cosθ=550,θ=Cos−1550

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $\bar{A}$ = 4 $\hat{i}$ - 2 $\hat{j}$ + 6 $\hat{k}$ , and $\bar{B}$ = $\hat{i}$ - 2 $\hat{j}$ - 3 $\hat{k}$ , the angle which the $\bar{A}$ + $\bar{B}$ makes with the x – axis is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\cos^{- 1} 3 / \sqrt{50}$

$\cos^{- 1} 4 / \sqrt{50}$

$\cos^{- 1} 5 / \sqrt{50}$

$\cos^{- 1} 12 / \sqrt{50}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \vec{A} + \vec{B} = 5 \hat{i} - 4 \hat{j} + 3 \hat{k} \\ (\vec{A} + \vec{B}) . \hat{i} = |\vec{A} + \vec{B}| . |\hat{i}| . Cos θ \\ \Rightarrow 5 = \sqrt{5^{2} + {(- 4)}^{2} + 3^{2}} .1. Cos θ \\ \Rightarrow Cos θ = \frac{5}{\sqrt{50}}, θ = {Cos}^{- 1} (\frac{5}{\sqrt{50}}) \end{array}$