If a- b- are unit vectors such that the vector a-3 b- is perpendicular to 7 a- 5b- and a- 4 b- is perpendicular to 7 a- -2b-then the angle between a- and b- is

We have (a⃗+3 b⃗)⊥ (7a⃗ 5b⃗) ⇒ (a⃗+3 b⃗).(7a⃗ 5b⃗)=0 ⇒ 7|a⃗|^2+16(a⃗.b⃗) 15|b⃗|^2=0 ⇒ 7+16 cos θ 15=0 ⇒ cos θ =1/2⇒θ =π/3 and (a⃗ 4b⃗)⊥ (7a⃗ 2b⃗) ⇒ (a⃗ 4b⃗).(7 ...

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Byju's Answer

Standard XII

Mathematics

Angle between Two Vectors

If a⃗, b⃗ are...

Question

If $\to a, \to b$ are unit vectors such that the vector $\to a + 3 \to b$ is perpendicular to $7 \to a - 5 \to b$ and $\to a - 4 \to b$ is perpendicular to $7 \to a - 2 \to b$ ,then the angle between $\to a$ and $\to b$ is

\frac{π}{6}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is C $\frac{π}{3}$
We have
$(\to a + 3 \to b) ⊥ (7 \to a - 5 \to b) \Rightarrow (\to a + 3 \to b) . (7 \to a - 5 \to b) = 0 \Rightarrow 7 | \to a |^{2} + 16 (\to a . \to b) - 15 | \to b |^{2} = 0 \Rightarrow 7 + 16 c o s θ - 15 = 0 \Rightarrow c o s θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$
and $(\to a - 4 \to b) ⊥ (7 \to a - 2 \to b)$
$\Rightarrow (\to a - 4 \to b) . (7 \to a - 2 \to b) = 0 \Rightarrow 7 | \to a |^{2} + 8 | \to b |^{2} - 30 (\to a . \to b) = 0 \Rightarrow 15 - 30 c o s θ = 0 \Rightarrow c o s θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$

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If →a,→b are unit vectors such that the vector →a+3→b is perpendicular to 7→a−5→b and →a−4→b is perpendicular to 7→a−2→b,then the angle between →a and →b is

If $\to a, \to b$ are unit vectors such that the vector $\to a + 3 \to b$ is perpendicular to $7 \to a - 5 \to b$ and $\to a - 4 \to b$ is perpendicular to $7 \to a - 2 \to b$ ,then the angle between $\to a$ and $\to b$ is