Let a and b be two vectors such that -2a- 3b- - -3a - b- and the angle between a and b is 60- If 18a is a unit vector- then -b- is equal to-

Given: nbsp;18a is a unit vector ⇒18|a| = 1 ⇒|a| = 8 Now, we know |2a+ 3b| = |3a + b| ⇒ nbsp;|2a+ 3b|^2= |3a + b|^2 ⇒ nbsp;4|a|^2 + 9|b|^2 + 12|a·b|= 9|a| ...

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Byju's Answer

Standard XII

Mathematics

Vector Triple Product

Let a and b b...

Question

Let $\to a$ and $\to b$ be two vectors such that $∣ ∣ ∣ 2 \to a + 3 \to b ∣ ∣ ∣ = ∣ ∣ ∣ 3 \to a + \to b ∣ ∣ ∣$ and the angle between $\to a$ and $\to b$ is $60^{\circ} .$ If $\frac{1}{8} \to a$ is a unit vector, then $∣ ∣ ∣ \to b ∣ ∣ ∣$ is equal to:

8

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Solution

The correct option is C $5$
Given: $\frac{1}{8} \to a$ is a unit vector
$\Rightarrow \frac{1}{8} ∣ ∣ \to a ∣ ∣ = 1 \Rightarrow ∣ ∣ \to a ∣ ∣ = 8$
Now, we know
$∣ ∣ ∣ 2 \to a + 3 \to b ∣ ∣ ∣ = ∣ ∣ ∣ 3 \to a + \to b ∣ ∣ ∣ \Rightarrow {∣ ∣ ∣ 2 \to a + 3 \to b ∣ ∣ ∣}^{2} = {∣ ∣ ∣ 3 \to a + \to b ∣ ∣ ∣}^{2} \Rightarrow 4 {∣ ∣ \to a ∣ ∣}^{2} + 9 {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + 12 ∣ ∣ ∣ \to a \cdot \to b ∣ ∣ ∣ = 9 {∣ ∣ \to a ∣ ∣}^{2} + {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + 6 ∣ ∣ ∣ \to a \cdot \to b ∣ ∣ ∣ \Rightarrow 8 {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + 6 ∣ ∣ ∣ \to a \cdot \to b ∣ ∣ ∣ - 5 {∣ ∣ \to a ∣ ∣}^{2} = 0 \Rightarrow 8 {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + 6 ∣ ∣ ∣ | \to a | \cdot | \to b | cos 60^{\circ} ∣ ∣ ∣ - 5 \times 64 = 0 \Rightarrow 8 {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + 24 ∣ ∣ ∣ \to b ∣ ∣ ∣ - 5 \times 64 = 0 \Rightarrow {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + 3 ∣ ∣ ∣ \to b ∣ ∣ ∣ - 40 = 0 \Rightarrow (| \to b | + 8) (| \to b | - 5) = 0 ∴ ∣ ∣ ∣ \to b ∣ ∣ ∣ = 5 [∵ ∣ ∣ ∣ \to b ∣ ∣ ∣ \geq 0]$

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Let →a and →b be two vectors such that ∣∣∣2→a+3→b∣∣∣=∣∣∣3→a+→b∣∣∣ and the angle between →a and →b is 60∘. If 18→a is a unit vector, then ∣∣∣→b∣∣∣ is equal to:

Let $\to a$ and $\to b$ be two vectors such that $∣ ∣ ∣ 2 \to a + 3 \to b ∣ ∣ ∣ = ∣ ∣ ∣ 3 \to a + \to b ∣ ∣ ∣$ and the angle between $\to a$ and $\to b$ is $60^{\circ} .$ If $\frac{1}{8} \to a$ is a unit vector, then $∣ ∣ ∣ \to b ∣ ∣ ∣$ is equal to: