Given vector A -2 - -3 - - the angle between A and y-axis is

Explanation for the correct option:Finding the angle:Given, A =2 î +3 ĵ0ex0ex|A|=√(2^2+3^2)0ex0ex =√(4+9)0ex0ex=√(13)0ex0ex𝑐𝑜𝑠(β)=𝐲/𝐚0ex0ex ...

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Standard XII

Mathematics

Addition of Vectors

Given vector ...

Question

$G i v e n v e c t o r \vec{A} = 2 \hat{i} + 3 \hat{j}, t h e a n g l e b e t w e e n \vec{A} a n d y - a x i s i s$

$\tan^{- 1} (3 / 2)$

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$\tan^{- 1} (2 / 3)$

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$\sin^{- 1} (2 / 3)$

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$\cos^{- 1} (2 / 3)$

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Solution

The correct option is B
$\tan^{- 1} (2 / 3)$

Explanation for the correct option:
Finding the angle:
$G i v e n, \vec{A} = 2 \hat{i} + 3 \hat{j} |A| = \sqrt{2^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13} c o s (β) = \frac{y}{a} = \frac{3}{\sqrt{13}} s i n (β) = \frac{x}{a} = \frac{2}{\sqrt{13}} t a n (β) = \frac{x}{y} = \frac{2}{3} β = t a n^{- 1} (\frac{2}{3})$
Hence form the given options option(B) is correct.

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GivenvectorA→=2i^+3j^,theanglebetweenA→andy-axisis

The correct option is B tan-1(2/3)Explanation for the correct option:Finding the angle:Given,A→=2i^+3j^A=22+32=4+9=13cos(β)=ya=313sin(β)=xa=213tan(β)=xy=23β=tan-1(23)Hence form the given options option(B) is correct.

$G i v e n v e c t o r \vec{A} = 2 \hat{i} + 3 \hat{j}, t h e a n g l e b e t w e e n \vec{A} a n d y - a x i s i s$