SATHEE: 3D Geometry 3 Question 20

3D Geometry 3 Question 20

####20. The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2 x + 3 y - z = 5$ , contains which one of the following points?

(2019 Main, 11 Jan I)

(a) $(- 2, 2, 2)$

(b) $(2, 2, 0)$

(d) $(0, - 2, 2)$

Show Answer

Answer:

(c)

Solution:

Let the direction vector of the line

$\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z - 1}{3} is b = 2 \hat{i} - \hat{j} + 3 \hat{k} .$

Since, the required plane contains this line and its projection along the plane $2 x + 3 y - z = 5$ , it will also contain the normal of the plane $2 x + 3 y - z = 5$ .

Normal vector of the plane $2 x + 3 y - z = 5$ is $n = 2 \hat{i} + 3 \hat{j} - \hat{k}$ .

Now, the required plane contains $b = 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $n = 2 \hat{i} + 3 \hat{j} - \hat{k}$ .

$∴$ Normal of the required plane is $b \times n$ .

Since, the plane contains the line $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z - 1}{3}$ , therefore it also contains the point $a = 3 \hat{i} - 2 \hat{j} + \hat{k}$ .

Now, the equation of required plane is $(r - a) \cdot (b \times n) = 0$

$| \begin{array}{ccc} x - 3 & y + 2 & z - 1 \\ 2 & - 1 & 3 \\ 2 & 3 & - 1 \end{array} | = 0$

$\Rightarrow (x - 3) [1 - 9] - (y + 2) [- 2 - 6] + (z - 1) [6 + 2] = 0$

$\Rightarrow - 8 x + 8 y + 8 z + 32 = 0$

$\Rightarrow x - y - z = 4$

Note that $(2, 0, - 2)$ is the only point which satisfy above equation.

3D Geometry 3 Question 20

Answer:

Solution:

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3D Geometry 3 Question 20

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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