Given that a and b are two unit vectors such that the angle between a and b is cos-1 14 - Let c be a vector in the plane of a and b- such that -c-4 and c-b - 2 a-b- If c - -a -b- where - and - are constants- then which of the following is-are CORRECT-

Given c×b = 2a×b ⇒ (c 2a) ×b =0 So, (c 2a) is parallel to b nbsp; ⇒c 2a = μb ⇒c =2a+μb ⋯ (1) Comparing withc=λa+ μb, we get λ nbsp; =2 Now, |a|=|b|=1, ...

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Byju's Answer

Standard XII

Mathematics

Dot Product of Two Vectors

Given that a ...

Question

Given that $\to a$ and $\to b$ are two unit vectors such that the angle between $\to a$ and $\to b$ is ${cos}^{- 1} (\frac{1}{4}) .$ Let $\to c$ be a vector in the plane of $\to a$ and $\to b,$ such that $| \to c | = 4$ and $\to c \times \to b = 2 \to a \times \to b .$ If $\to c = λ \to a + μ \to b,$ where $λ$ and $μ$ are constants, then which of the following is/are CORRECT?

λ = 2.

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Sum of all possible values of

μ

- 1.

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Product of all possible values of

μ

- 12.

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Number of distinct values of

μ

3.

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Solution

The correct option is C Product of all possible values of $μ$ is $- 12.$
Given $\to c \times \to b = 2 \to a \times \to b$
$\Rightarrow (\to c - 2 \to a) \times \to b = \to 0$

So, $(\to c - 2 \to a)$ is parallel to $\to b$
$\Rightarrow \to c - 2 \to a = μ \to b \Rightarrow \to c = 2 \to a + μ \to b \dots (1)$
Comparing with $\to c = λ \to a + μ \to b$ , we get
$λ = 2$

Now, $| \to a | = | \to b | = 1, \to a \cdot \to b = \frac{1}{4}$
Taking dot product with $\to a, \to b$ and $\to c$ simultaneously with equation $(1)$ , we get
$\to a \cdot \to c = 2 + \frac{μ}{4} \dots (2) \to b \cdot \to c = \frac{1}{2} + μ \dots (3) 16 = 2 \to a \cdot \to c + μ \to b \cdot \to c$
Using equation $(2)$ and $(3)$ , we get
$16 = 4 + \frac{μ}{2} + \frac{μ}{2} + μ^{2} \Rightarrow μ^{2} + μ - 12 = 0 \Rightarrow (μ + 4) (μ - 3) = 0 \Rightarrow μ = - 4, 3$

Hence, the sum of all possible values of $μ$ is $- 1.$
Product of all possible values of $μ$ is $- 12.$
Number of distinct values of $μ$ is $2.$

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