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AP DSC School Assistant Math Model Test 3

Section: Subject Specific

Question : 43 of 160

Marks: +1, -0

(\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b})=27

| \vec{a} | =2 | \vec{b} |

| \vec{b} |

is :

Solution:

Concept:

The dot product of two vectors expands using distributive and commutative properties:

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

. Also,

|\vec{a}| = 2|\vec{b}|

implies

|\vec{a}|^2 = 4|\vec{b}|^2

Explanation:

Let

|\vec{b}| = x

. Then

|\vec{a}| = 2x

, so

|\vec{a}|^2 = 4x^2

and

|\vec{b}|^2 = x^2

From the given condition:

(\vec{a}-\vec{b})\cdot(\vec{a}+\vec{b}) = 27

Using the identity:

27 = |\vec{a}|^2 - |\vec{b}|^2 = 4x^2 - x^2 = 3x^2

Thus,

3x^2 = 27

x^2 = 9

x = 3

(since magnitude is positive).

Answer:

|\vec{b}| = 3

units, which corresponds to option A.

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