If A -4 -2 -6 k- and B -2 -3 k- the angle which A - B makes with x-axis isA- cos -11-10B- cos -12-10cos -11-5cos -11-10

The correct option is D cos^ 1(1√(10))A = 4î 2ĵ + 6k̂ amp; B = nbsp;î 2ĵ 3k̂ A B = 3î+9k̂ Let x axis = î cos θ = (3î+9k̂) . (î)√(3^2+9^2) cos θ = 3√ ...

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Byju's Answer

Standard XII

Physics

Vector Component

If A =4 î-2 ĵ...

Question

If $\to A = 4^i - 2^j + 6^k$ and $\to B =^i - 2^j - 3^k$ the angle which $\to A - \to B$ makes with x-axis is

{cos}^{- 1} (\frac{1}{10})

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{cos}^{- 1} (\frac{2}{\sqrt{10}})

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{cos}^{- 1} (\frac{1}{\sqrt{10}})

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{cos}^{- 1} (\frac{1}{5})

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Solution

The correct option is C ${cos}^{- 1} (\frac{1}{\sqrt{10}})$
$\to A = 4^i - 2^j + 6^k$ & $\to B =^i - 2^j - 3^k$
$\to A - \to B = 3^i + 9^k$
Let x - axis = $^i$
$cos θ = \frac{(3^i + 9^k) . (^i)}{\sqrt{3^{2} + 9^{2}}}$
$cos θ = \frac{3}{\sqrt{90}}$
$θ = {cos}^{- 1} \frac{1}{\sqrt{10}}$

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

The correct option is C cos−1(1√10)→A=4^i−2^j+6^k & →B=^i−2^j−3^k →A−→B=3^i+9^k Let x - axis = ^i cos θ=(3^i+9^k).(^i)√32+92 cos θ=3√90 θ=cos−11√10

If $\to A = 4^i - 2^j + 6^k$ and $\to B =^i - 2^j - 3^k$ the angle which $\to A - \to B$ makes with x-axis is

The correct option is C ${cos}^{- 1} (\frac{1}{\sqrt{10}})$
$\to A = 4^i - 2^j + 6^k$ & $\to B =^i - 2^j - 3^k$
$\to A - \to B = 3^i + 9^k$
Let x - axis = $^i$
$cos θ = \frac{(3^i + 9^k) . (^i)}{\sqrt{3^{2} + 9^{2}}}$
$cos θ = \frac{3}{\sqrt{90}}$
$θ = {cos}^{- 1} \frac{1}{\sqrt{10}}$