Let -a and -bbe two non-zero vectors perpendicular to each other and-a-b- If -a-b-a- then the angle between the vectors -a-b-a-b and -a is equal to-

Step 1: Determine the magnitude of given vectorsLet ⇀a and ⇀b be two non zero vectors perpendicular to each other.|⇀a×⇀b|=|⇀a| |⇀b| sin(90)°0ex0ex⇒|⇀a|=|⇀a| |⇀a ...

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Byju's Answer

Standard VII

Mathematics

Euler's Formula

Let ⇀a and ⇀b...

Question

Let $\overset{⇀}{a}$ and $\overset{⇀}{b}$ be two non-zero vectors perpendicular to each other and $|\overset{⇀}{a}| = |\overset{⇀}{b}|$ . If $|\overset{⇀}{a} \times \overset{⇀}{b}| = |\overset{⇀}{a}|$ , then the angle between the vectors $(\overset{⇀}{a} + \overset{⇀}{b} + (\overset{⇀}{a} \times \overset{⇀}{b}))$ and $\overset{⇀}{a}$ is equal to.

$\sin^{- 1} (\frac{1}{\sqrt{3}})$

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$\cos^{- 1} (\frac{1}{\sqrt{3}})$

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$\sin^{- 1} (\frac{1}{\sqrt{6}})$

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

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Solution

The correct option is B
$\cos^{- 1} (\frac{1}{\sqrt{3}})$

Step 1: Determine the magnitude of given vectors
Let $\overset{⇀}{a}$ and $\overset{⇀}{b}$ be two non-zero vectors perpendicular to each other.
$|\overset{⇀}{a} \times \overset{⇀}{b}| = |\overset{⇀}{a}| |\overset{⇀}{b}| \sin (90) ° \Rightarrow |\overset{⇀}{a}| = |\overset{⇀}{a}| |\overset{⇀}{a}| 1 [∵ |\overset{⇀}{a}| = |\overset{⇀}{b}|, |\overset{⇀}{a} \times \overset{⇀}{b}| = |\overset{⇀}{a}|] \Rightarrow |\overset{⇀}{a}| = 1 \Rightarrow |\overset{⇀}{b}| = 1; |\overset{⇀}{a} \times \overset{⇀}{b}| = 1$
Therefore $\overset{⇀}{a}$ , $\overset{⇀}{b}$ and $|\overset{⇀}{a} \times \overset{⇀}{b}|$ are mutually perpendicular unit vectors.
We can take, $\hat{a} = \overset{⇀}{i}; \hat{b} = \overset{⇀}{j}; \hat{c} = \overset{⇀}{k}$
Step 2: Determine the angle between the vectors
The angle between $(\overset{⇀}{a} + \overset{⇀}{b} + (\overset{⇀}{a} \times \overset{⇀}{b}))$ and $\overset{⇀}{a}$ , is given by using dot product
$(\overset{⇀}{a} + \overset{⇀}{b} + (\overset{⇀}{a} \times \overset{⇀}{b})) . \overset{⇀}{a} = |\overset{⇀}{a} + \overset{⇀}{b} + (\overset{⇀}{a} \times \overset{⇀}{b})| |\overset{⇀}{a}| \cos (θ) (\overset{⇀}{i} + \overset{⇀}{j} + \overset{⇀}{k}) . \overset{⇀}{i} = |\overset{⇀}{i} + \overset{⇀}{j} + \overset{⇀}{k}| |\overset{⇀}{i}| \cos (θ) 1 = (\sqrt{1^{2} + 1^{2} + 1^{2}}) (\sqrt{1^{2}}) \cos (θ) \cos (θ) = \frac{1}{\sqrt{3}} θ = \cos^{- 1} (\frac{1}{\sqrt{3}})$
Hence, option (B) is the correct answer.

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Let a⇀ and b⇀be two non-zero vectors perpendicular to each other anda⇀=b⇀. If a⇀×b⇀=a⇀, then the angle between the vectors a⇀+b⇀+a⇀×b⇀ and a⇀ is equal to.