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Question

Answers

(A) $\dfrac{3}{4}$

(B) $\dfrac{2}{3}$

(C) $-\dfrac{5}{2}$

(D) $-\dfrac{2}{3}$

Answer
Verified

Hint: Two vectors are perpendicular if the angle between them is $\dfrac{\pi }{2}$ , this means that their dot product must be $0$ as ${cos \left(\dfrac{\pi }{2}\right)\ }=0$. The value of $\mu$ can be found by using $(a+\mu b).(a-\mu b)=0$ where $a,b$ are the given vectors with known magnitudes.

Complete step-by-step answer:

Since we want the vectors $(a+\mu b)$ and $(a-\mu b)$ to be perpendicular , their dot product must be $0$. The dot product of two vectors ~$\overrightarrow{x}\ $and $\overrightarrow{y}$ is given by,

$$\overrightarrow{x}.\overrightarrow{y}=\left|\overrightarrow{x}\right|\left|\overrightarrow{y}\right|{cos \theta \ }$$

Here $\theta $ is the angle between the two vectors. Thus if the vectors are perpendicular then the angle between them is ${90}^{\circ }$ and ${cos {90}^{\circ }\ }=0$ is the factor which makes the dot product $0$.

The dot product of $(a+\mu b)$ and $(a-\mu b)$ can be simplified as follows ,

$$\left(a+\mu b\right).\left(a-\mu b\right)=a.a-\mu \left(a.b\right)+\mu \left(b.a\right)-{\mu }^2\left(b.b\right)$$

Since by rules of dot product we know that $a.a={\left|a\right|}^2$ and that dot product is commutative which means $a.b=b.a$ . This gives us,

$$\left(a+\mu b\right)\left(a-\mu b\right)={|a|}^2-{\mu }^2|{b|}^2$$

Since the dot product is $0$ and $\left|a\right|=3,\left|b\right|=4$ we can conclude that ${\left|a\right|}^2-{\mu }^2{\left|b\right|}^2=0$ gives ,

$${\mu }^2=\dfrac{{\left|a\right|}^2}{{\left|b\right|}^2}=\dfrac{9}{16}$$

This gives two values of $\mu $ namely $\mu =\pm \dfrac{3}{4}$ . Option (A) suggests one of these answers and thus $\mu =\dfrac{3}{4}$ is the correct answer.

Note: Whenever we have perpendicular or parallel vectors, we must think of the scalar and vector products which vanish respectively if the vectors are perpendicular or parallel. An alternative way is to think it in terms of a right angled triangle with sides $(a+\mu b), (a-\mu b), (2\mu b)$ where we must have $|a+\mu b|^2+|a-\mu b|^2 = |2\mu b|^2$.

Complete step-by-step answer:

Since we want the vectors $(a+\mu b)$ and $(a-\mu b)$ to be perpendicular , their dot product must be $0$. The dot product of two vectors ~$\overrightarrow{x}\ $and $\overrightarrow{y}$ is given by,

$$\overrightarrow{x}.\overrightarrow{y}=\left|\overrightarrow{x}\right|\left|\overrightarrow{y}\right|{cos \theta \ }$$

Here $\theta $ is the angle between the two vectors. Thus if the vectors are perpendicular then the angle between them is ${90}^{\circ }$ and ${cos {90}^{\circ }\ }=0$ is the factor which makes the dot product $0$.

The dot product of $(a+\mu b)$ and $(a-\mu b)$ can be simplified as follows ,

$$\left(a+\mu b\right).\left(a-\mu b\right)=a.a-\mu \left(a.b\right)+\mu \left(b.a\right)-{\mu }^2\left(b.b\right)$$

Since by rules of dot product we know that $a.a={\left|a\right|}^2$ and that dot product is commutative which means $a.b=b.a$ . This gives us,

$$\left(a+\mu b\right)\left(a-\mu b\right)={|a|}^2-{\mu }^2|{b|}^2$$

Since the dot product is $0$ and $\left|a\right|=3,\left|b\right|=4$ we can conclude that ${\left|a\right|}^2-{\mu }^2{\left|b\right|}^2=0$ gives ,

$${\mu }^2=\dfrac{{\left|a\right|}^2}{{\left|b\right|}^2}=\dfrac{9}{16}$$

This gives two values of $\mu $ namely $\mu =\pm \dfrac{3}{4}$ . Option (A) suggests one of these answers and thus $\mu =\dfrac{3}{4}$ is the correct answer.

Note: Whenever we have perpendicular or parallel vectors, we must think of the scalar and vector products which vanish respectively if the vectors are perpendicular or parallel. An alternative way is to think it in terms of a right angled triangle with sides $(a+\mu b), (a-\mu b), (2\mu b)$ where we must have $|a+\mu b|^2+|a-\mu b|^2 = |2\mu b|^2$.

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