Let a- and c- be unit vectors and -b-4 with a-b-2 a-c- The angle between a- and c- is cos -11-4- If b-2 c -a- then - is equal toA- 1-3- 2-3B- 2-3C- 3-4 1-3- 1-2

The correct option is C 3, 4It is given that the angle between nbsp;a and c is cos^ 1(14). So, a·c=|a||c|cos(cos^ 1(14)) ⇒a·c=14 ⋯(1) b 2c=λa Taking dot ...

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Byju's Answer

Standard XII

Mathematics

Dot Product of Two Vectors

Let a⃗ and c⃗...

Question

Let $\to a$ and $\to c$ be unit vectors and $| \to b | = 4$ with $\to a \times \to b = 2 \to a \times \to c .$ The angle between $\to a$ and $\to c$ is ${cos}^{- 1} (\frac{1}{4}) .$ If $\to b - 2 \to c = λ \to a,$ then $λ$ is equal to

\frac{1}{3}, \frac{1}{2}

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\frac{1}{3}, \frac{2}{3}

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3, - 4

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2, 3

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Solution

The correct option is C $3, - 4$
It is given that the angle between $\to a$ and $\to c$ is ${cos}^{- 1} (\frac{1}{4}) .$
So, $\to a \cdot \to c = | \to a | | \to c | cos ({cos}^{- 1} (\frac{1}{4}))$
$\Rightarrow \to a \cdot \to c = \frac{1}{4} \dots (1)$

$\to b - 2 \to c = λ \to a$
Taking dot products with $\to a,$ we have
$\to a \cdot \to b - 2 (\to a \cdot \to c) = λ (\to a \cdot \to a)$
$\Rightarrow \to a \cdot \to b - \frac{1}{2} = λ$
$\Rightarrow \to a \cdot \to b = \frac{1}{2} + λ$

Similarly, taking dot products with $\to b$ and $\to c$
$\to b \cdot \to c = 8 - \frac{λ^{2}}{2} - \frac{λ}{4} \dots (2)$
and $\to b \cdot \to c - 2 = λ (\to a \cdot \to c) \dots (3)$

From equations $(1), (2)$ and $(3),$ we get
$8 - \frac{λ^{2}}{2} - \frac{λ}{4} - 2 = λ (\frac{1}{4})$
$\Rightarrow λ = 3, - 4$

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Let $\to a, \to b, \to c$ be three vectors such that $| \to a | = | \to c | = 1; | \to b | = 4$ and $| \to b \times \to c | = \sqrt{15}$ . If $\to b - 2 \to c = λ \to a$ then a value of $λ$ is