SATHEE: Vectors 1 Question 27

Vectors 1 Question 27

28. Determine the value of $c$ , so that for all real $x$ , the vector $c x \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $x \hat{i} + 2 \hat{j} + 2 c x \hat{k}$ make an obtuse angle with each other.

(1991, 4M)

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Answer:

Correct Answer: 28. (3)

Solution:

Let $\vec{a} = c x \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $\vec{b} = x \hat{i} + 2 \hat{j} + 2 c x \hat{k}$ . Since, $\vec{a}$ and $\vec{b}$ makes an obtuse angle.

$\begin{aligned} \Rightarrow \vec{a} \cdot \vec{b} < 0 \Rightarrow c x^{2} - 12 + 6 c x < 0 \\ \Rightarrow c < 0 and discriminant < 0 \\ \Rightarrow c < 0 and 36 c^{2} - 4 \cdot (- 12) c < 0 \\ \Rightarrow c < 0 and 12 c (3 c + 4) < 0 \\ \Rightarrow c < 0 and c > - 4 / 3 \\ ∴ c \in (- 4 / 3, 0) \end{aligned}$

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Vectors 1 Question 27

28. Determine the value of c, so that for all real x, the vector cxi^−6j^−3k^ and xi^+2j^+2cxk^ make an obtuse angle with each other.

Answer:

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28. Determine the value of $c$ , so that for all real $x$ , the vector $c x \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $x \hat{i} + 2 \hat{j} + 2 c x \hat{k}$ make an obtuse angle with each other.