SATHEE: Vectors 2 Question 3

Vectors 2 Question 3

3. If the length of the perpendicular from the point $(β, 0, β) (β \neq 0)$ to the line, $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{- 1}$ is $\sqrt{\frac{3}{2}}$ , then $β$ is equal to

(2019 Main, 10 April +I)

(a) 2

(b) -2

(d) 1

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Answer:

Correct Answer: 3. (c)

Solution:

Equation of given line is

$\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{- 1}$

Now, one of the point on line is $P (0, 1, - 1)$ and the given point is $Q (β, 0, β)$ .

From the figure, the length of the perpendicular

$\begin{array}{ll} Q M = l = \sqrt{\frac{3}{2}} \\ \Rightarrow & \frac{| P Q \times P M |}{| P M |} = \sqrt{\frac{3}{2}} \\ ∵ & P Q = β \hat{i} - \hat{j} + (β + 1) \hat{k} \end{array}$

and $P M =$ a vector along given line $(i) = \hat{i} - \hat{k}$ So, $P Q \times P M = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} β & - 1 & β + 1 1 & 0 & - 1 \end{array} |$

$= \hat{i} - \hat{j} (- β - β - 1) + \hat{k} = \hat{i} + (2 β + 1) \hat{j} + \hat{k}$

Now, $\frac{| P Q \times P M |}{| P M |} = \frac{\sqrt{1 + (2 β + 1)^{2} + 1}}{\sqrt{2}}$

From Eqs. (ii) and (iii), we get

$\begin{aligned} \sqrt{\frac{1 + (2 β + 1)^{2} + 1}{2}} = \sqrt{\frac{3}{2}} \Rightarrow \frac{1 + (2 β + 1)^{2} + 1}{2} = \frac{3}{2} \\ \Rightarrow & (2 β + 1)^{2} = 1 \Rightarrow 2 β + 1 = \pm 1 \\ \Rightarrow & 2 β + 1 = 1 or 2 β + 1 = - 1 \Rightarrow β = 0 or β = - 1 \end{aligned}$

Vectors 2 Question 3

3. If the length of the perpendicular from the point $(β, 0, β) (β \neq 0)$ to the line, $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{- 1}$ is $\sqrt{\frac{3}{2}}$ , then $β$ is equal to

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Vectors 2 Question 3

3. If the length of the perpendicular from the point (β,0,β)(β≠0) to the line, x1=y−10=z+1−1 is 32, then β is equal to

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3. If the length of the perpendicular from the point $(β, 0, β) (β \neq 0)$ to the line, $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{- 1}$ is $\sqrt{\frac{3}{2}}$ , then $β$ is equal to