SATHEE: 3D Geometry 3 Question 12

3D Geometry 3 Question 12

####12. The length of the perpendicular from the point $(2, - 1, 4)$ on the straight line, $\frac{x + 3}{10} = \frac{y - 2}{- 7} = \frac{z}{1}$ is

(a) greater than 3 but less than 4

(b) less than 2

(d) greater than 4

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Answer:

(a)

Solution:

Equation of given line is

$\frac{x + 3}{10} = \frac{y - 2}{- 7} = \frac{z}{1} = r (let)$

Coordinates of a point on line (i) is

$A (10 r - 3, - 7 r + 2, r)$

Now, let the line joining the points $P (2, - 1, 4)$ and $A (10 r - 3, - 7 r + 2, r)$ is perpendicular to line (i). Then,

$P A \cdot (10 \hat{i} - 7 \hat{j} + \hat{k}) = 0$

$[∵$ vector along line (i) is $(10 \hat{i} - 7 \hat{j} + \hat{k})]$

$\Rightarrow [(10 r - 5) \hat{i} + (- 7 r + 3) \hat{j} + (r - 4) \hat{k}] \cdot [10 \hat{i} - 7 \hat{j} + \hat{k}] = 0$

$\Rightarrow 10 (10 r - 5) - 7 (3 - 7 r) + (r - 4) = 0$

$\Rightarrow 100 r - 50 - 21 + 49 r + r - 4 = 0$

$\Rightarrow 150 r = 75 \Rightarrow r = \frac{1}{2}$

So, the foot of perpendicular is $A 2, - \frac{3}{2}, \frac{1}{2}$

[put $r = \frac{1}{2}$ in the coordinates of point $A$ ]

Now, perpendicular distance of point $P (2, - 1, 4)$ from the line (i) is

$\begin{aligned} P A & = \sqrt{(2 - 2)^{2} + - \frac{3}{2} + 1^{2} + \frac{1}{2} - 4^{2}} & = \sqrt{\frac{1}{4} + \frac{49}{4}} = \sqrt{\frac{50}{4}} = \frac{5}{\sqrt{2}} \end{aligned}$

which lies in $(3, 4)$ .

3D Geometry 3 Question 12

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3D Geometry 3 Question 12

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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