In the angle subtended by vector i - j with the x - axis is-

The correct option is B If a̅ = î + ĵ; b̅ = î (x axis) then the angle between a̅ amp; b̅ is cos θ = a̅.b̅/|a̅| |b̅| = (î + ĵ).(î)/√(1^2 + 1^2)×√(1^2) =1/√(2 ...

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Standard XII

Physics

Unit Vectors

In the angle ...

Question

In the angle subtended by vector $ˆ i + ˆ j$ with the x - axis is

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Solution

The correct option is B
If $¯ a =^i +^j; ¯ b =^i$ (x-axis)
then the angle between $¯ a$ & $¯ b$ is cos $θ = \frac{¯ a . ¯ b}{| ¯ a | | ¯ b |} = \frac{(^i +^j) . (^i)}{\sqrt{1^{2} + 1^{2}} \times \sqrt{1^{2}}} = \frac{1}{\sqrt{2}}$
$\Rightarrow θ = 45^{o}$

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In the angle subtended by vector ˆi+ˆj with the x - axis is

The correct option is B If ¯a=^i+^j;¯b=^i (x-axis) then the angle between ¯a & ¯b is cos θ=¯a.¯b|¯a| |¯b|=(^i+^j).(^i)√12+12×√12=1√2 ⇒θ=45o

In the angle subtended by vector $ˆ i + ˆ j$ with the x - axis is

The correct option is B
If $¯ a =^i +^j; ¯ b =^i$ (x-axis)
then the angle between $¯ a$ & $¯ b$ is cos $θ = \frac{¯ a . ¯ b}{| ¯ a | | ¯ b |} = \frac{(^i +^j) . (^i)}{\sqrt{1^{2} + 1^{2}} \times \sqrt{1^{2}}} = \frac{1}{\sqrt{2}}$
$\Rightarrow θ = 45^{o}$