Let a and b be two vectors such that | a| = 1, | b| = 4 and a b = 2. If c = (2 a b) - 3 b and the angle between b and c is, then 192 ^2() is equal to:

Given: | a| = 1, | b| = 4, a b = 2 and c = (2 a b) - 3 b. Find: 192 ^2, where is the angle between b and c. Using dot product, b c = (2 a b) b - 3| b|^2 Now ( a b) b = 0, so b c = -3| b|^2 = -3 16 = -48 Also, b c = | b| | c| Hence, 4| c| = -48 so | c| = -12 Now from a b = | a| | b|, 2 = 1 4 therefore = 12 and so ^2 = 1 - 14 = 34 Next, | c|^2 = |(2 a b) - 3 b|^2 From the extracted working, | c|^2 = 64 34 + 144 = 192 Also, | c|^2 ^2 = 144 Thus, 192 ^2 = 144 Therefore, 192 ^2 = 192(1- ^2) = 192 - 144 = 48 So, the correct option is A. Given: c = (2 a b) - 3 b. Find: 192 ^2. First use b c = -3| b|^2 = -48 So, | b| | c| = -48 With | b|=4, | c| = -12 Hence, | c|^2 ^2 = 144 From the solution working, | c|^2 = 192 Therefore, 192 ^2 = 144 which gives 192 ^2 = 192 - 144 = 48 Therefore, the required value is 48.

Let a and b be two vectors such that | a| = 1, | b| = 4 and a b =… — Solution | JEE Mathematics

MCQMediumJEE 2024Cross Product

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}| = 1$ , $|\vec{b}| = 4$ and $\vec{a} \cdot \vec{b} = 2$ . If $\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $\alpha$ , then $192 \sin^2(\alpha)$ is equal to:

A
$48$
B
$54$
C
$60$
D
$72$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $|\vec{a}| = 1$ , $|\vec{b}| = 4$ , $\vec{a} \cdot \vec{b} = 2$ and $\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$ .

Find: $192\sin^2\alpha$ , where $\alpha$ is the angle between $\vec{b}$ and $\vec{c}$ .

Using dot product,

\vec{b} \cdot \vec{c} = (2\vec{a} \times \vec{b}) \cdot \vec{b} - 3|\vec{b}|^2

Now $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$ , so

\vec{b} \cdot \vec{c} = -3|\vec{b}|^2 = -3 \times 16 = -48

Also,

\vec{b} \cdot \vec{c} = |\vec{b}|\,|\vec{c}|\cos\alpha

Hence,

4|\vec{c}|\cos\alpha = -48

|\vec{c}|\cos\alpha = -12

Now from $\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta$ ,

2 = 1 \times 4 \cos\theta

therefore

\cos\theta = \frac{1}{2}

and so

\sin^2\theta = 1 - \frac{1}{4} = \frac{3}{4}

Next,

|\vec{c}|^2 = |(2\vec{a} \times \vec{b}) - 3\vec{b}|^2

From the extracted working,

|\vec{c}|^2 = 64 \times \frac{3}{4} + 144 = 192

Also,

|\vec{c}|^2 \cos^2\alpha = 144

Thus,

192\cos^2\alpha = 144

Therefore,

192\sin^2\alpha = 192(1-\cos^2\alpha) = 192 - 144 = 48

So, the correct option is A.

Direct Relation

Given: $\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$ .

Find: $192\sin^2\alpha$ .

First use

\vec{b} \cdot \vec{c} = -3|\vec{b}|^2 = -48

So,

|\vec{b}|\,|\vec{c}|\cos\alpha = -48

With $|\vec{b}|=4$ ,

|\vec{c}|\cos\alpha = -12

Hence,

|\vec{c}|^2\cos^2\alpha = 144

From the solution working,

|\vec{c}|^2 = 192

Therefore,

192\cos^2\alpha = 144

which gives

192\sin^2\alpha = 192 - 144 = 48

Therefore, the required value is $48$ .

Common mistakes

Using $\vec{b} \cdot (\vec{a} \times \vec{b})$ as a non-zero term. This is wrong because $\vec{a} \times \vec{b}$ is perpendicular to $\vec{b}$ . Therefore this dot product is zero, and only the $-3\vec{b}$ term contributes.
Finding the angle between $\vec{a}$ and $\vec{b}$ and treating it as $\alpha$ . This is wrong because $\alpha$ is the angle between $\vec{b}$ and $\vec{c}$ . First compute the relation involving $\vec{b} \cdot \vec{c}$ , then use it for $\alpha$ .
Computing $192\sin^2\alpha$ directly from $\cos\alpha$ without using $|\vec{c}|^2$ . This is incomplete because the relation obtained is for $|\vec{c}|\cos\alpha$ , not for $\cos\alpha$ alone. One must first determine $|\vec{c}|^2$ .

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Answer

Step-by-step solution

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