Let a and b be two vectors such that | a| = 14, | b| = 6, | a b| = 48. Then ( a b)^2 is equal to:

Given: | a| = 14, | b| = 6, and | a b| = 48. Find: The value of ( a b)^2. Use the identity | a b|^2 + ( a b)^2 = | a|^2 | b|^2 Substitute the given values: ( 48)^2 + ( a b)^2 = ( 14)^2( 6)^2 So, 48 + ( a b)^2 = 14 6 = 84 Therefore, ( a b)^2 = 84 - 48 = 36 Therefore, the required value is 36. Given: | a| = 14, | b| = 6, | a b| = 48. Find: ( a b)^2. The required identity is | a b|^2 + ( a b)^2 = | a|^2 | b|^2 Now compute each squared magnitude: | a b|^2 = ( 48)^2 = 48 | a|^2 = ( 14)^2 = 14 | b|^2 = ( 6)^2 = 6 Substitute into the identity: 48 + ( a b)^2 = 14 6 48 + ( a b)^2 = 84 Hence, ( a b)^2 = 84 - 48 = 36 So, the correct numerical value is 36.

Let a and b be two vectors such that | a| = 14, | b| = 6, | a b|… — Solution | JEE Mathematics

NVAEasyJEE 2023Cross Product

Let $\vec{a}$ and $\vec{b}$ be two vectors such that

|\vec{a}| = \sqrt{14}, \quad |\vec{b}| = \sqrt{6}, \quad |\vec{a} \times \vec{b}| = \sqrt{48}.

Then $(\vec{a} \cdot \vec{b})^2$ is equal to:

Answer

Correct answer:36

Step-by-step solution

Standard Method

Given: $|\vec{a}| = \sqrt{14}$ , $|\vec{b}| = \sqrt{6}$ , and $|\vec{a} \times \vec{b}| = \sqrt{48}$ .

Find: The value of $(\vec{a} \cdot \vec{b})^2$ .

Use the identity

|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2

Substitute the given values:

(\sqrt{48})^2 + (\vec{a} \cdot \vec{b})^2 = (\sqrt{14})^2(\sqrt{6})^2

So,

48 + (\vec{a} \cdot \vec{b})^2 = 14 \times 6 = 84

Therefore,

(\vec{a} \cdot \vec{b})^2 = 84 - 48 = 36

Therefore, the required value is $36$ .

Step-by-step Calculation

Given: $|\vec{a}| = \sqrt{14}$ , $|\vec{b}| = \sqrt{6}$ , $|\vec{a} \times \vec{b}| = \sqrt{48}$ .

Find: $(\vec{a} \cdot \vec{b})^2$ .

The required identity is

|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2

Now compute each squared magnitude:

|\vec{a} \times \vec{b}|^2 = (\sqrt{48})^2 = 48

|\vec{a}|^2 = (\sqrt{14})^2 = 14

|\vec{b}|^2 = (\sqrt{6})^2 = 6

Substitute into the identity:

48 + (\vec{a} \cdot \vec{b})^2 = 14 \times 6

48 + (\vec{a} \cdot \vec{b})^2 = 84

Hence,

(\vec{a} \cdot \vec{b})^2 = 84 - 48 = 36

So, the correct numerical value is $36$ .

Common mistakes

Using $|\vec{a} \times \vec{b}|$ instead of $|\vec{a} \times \vec{b}|^2$ in the identity is incorrect because the relation involves squared quantities. First square the given cross product magnitude, then substitute.
Confusing $|\vec{a}|^2 |\vec{b}|^2$ with $|\vec{a}| |\vec{b}|$ is wrong. Here, $|\vec{a}| = \sqrt{14}$ and $|\vec{b}| = \sqrt{6}$ , so the product on the right side becomes $14 \times 6$ after squaring.
Taking $\vec{a} \cdot \vec{b} = 36$ instead of $(\vec{a} \cdot \vec{b})^2 = 36$ is incorrect. The question asks for the square of the dot product, not the dot product itself.

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