SATHEE: 3D Geometry 3 Question 21

3D Geometry 3 Question 21

####21. The direction ratios of normal to the plane through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π / 4$ with the plane $y - z + 5 = 0$ are

(2019 Main, 11 Jan I)

(a) $2, - 1, 1$

(b) $\sqrt{2}, 1, - 1$

(d) $2 \sqrt{3}, 1, - 1$

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Answer:

(b,c)

Solution:

Let the equation of plane be

$a (x - 0) + b (y + 1) + c (z - 0) = 0$

$[∵$ Equation of plane passing through a point $(x_{1}, y_{1}, z_{1})$ is given by $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0]$ $\Rightarrow a x + b y + c z + b = 0$

Since, it also passes through $(0, 0, 1)$ , therefore, we get

$c + b = 0$

Now, as angle between the planes

$a x + b y + c z + b = 0$

and

$y - z + 5 = 0 is \frac{π}{4}$

$∴ \cos \frac{π}{4} = \frac{| n_{1} \cdot n_{2} |}{| n_{1} | | n_{2} |};$ where $n_{1} = a \hat{i} + b \hat{j} + c \hat{k}$

and $n_{2} = 0 \hat{i} + \hat{j} - \hat{k}$

$\begin{aligned} \Rightarrow \frac{1}{\sqrt{2}} & = \frac{| (a \hat{i} + b \hat{j} + c \hat{k}) \cdot (0 \hat{i} + \hat{j} - \hat{k}) |}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{0 + 1 + 1}} \\ = \frac{| b - c |}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{2}} \end{aligned}$

$\Rightarrow a^{2} + b^{2} + c^{2} = | b - c |^{2} = (b - c)^{2} = b^{2} + c^{2} - 2 b c$

$\Rightarrow a^{2} = - 2 b c$

$\Rightarrow a^{2} = 2 b^{2}$

$\Rightarrow a = \pm \sqrt{2} b$

[Using Eq. (ii)]

$\Rightarrow$ Direction ratios $(a, b, c) = (\pm \sqrt{2}, 1, - 1)$

So, options (b) and (c) are correct because

$2, \sqrt{2}, - \sqrt{2}$ and $\sqrt{2}, 1, - 1$ . are multiple of each other.

3D Geometry 3 Question 21

Answer:

Solution:

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3D Geometry 3 Question 21

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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