SATHEE: 3D Geometry 3 Question 24

3D Geometry 3 Question 24

24. The plane passing through the point $(4, - 1, 2)$ and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point

(a) $(- 1, - 1, - 1)$

(b) $(1, 1, - 1)$

(d) $(- 1, - 1, 1)$

(2019 Main, 10 Jan I)

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Answer:

(c)

Solution:

Let $a$ be the position vector of the given point $(4, - 1, 2)$ .

Then, $a = 4 \hat{i} - \hat{j} + 2 \hat{k}$

The direction vector of the lines

$\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$

are respectively

$\begin{array}{r} b_{1} = 3 \hat{i} - \hat{j} + 2 \hat{k} \\ and b_{2} = \hat{i} + 2 \hat{j} + 3 \hat{k} \end{array}$

Now, as the plane is parallel to both $b_{1}$ and $b_{2}$

$[∵$ plane is parallel to the given lines]

So, normal vector (n) of the plane is perpendicular to both $b_{1}$ and $b_{2}$ .

$\Rightarrow n = b_{1} \times b_{2}$ and

Required equation of plane is

$\begin{aligned} (r - a) \cdot n = 0 \\ \Rightarrow (r - a) \cdot (b_{1} \times b_{2}) = 0 \\ \Rightarrow | \begin{array}{ccc} x - 4 & y + 1 & z - 2 \\ 3 & - 1 & 2 \\ 1 & 2 & 3 \end{array} | = 0 \\ ∵ r - a = (x \hat{i} + y \hat{j} + z \hat{k}) - (4 \hat{i} - \hat{j} + 2 \hat{k}) \\ = (x - 4) \hat{i} + (y + 1) \hat{j} + (z - 2) \hat{k} \end{aligned}$

and we know that, $[a b c] = a \cdot (b \times c)$

$= | \begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array} |$

$\Rightarrow (x - 4) (- 3 - 4) - (y + 1) (9 - 2) + (z - 2) (6 + 1) = 0$

$\Rightarrow - 7 (x - 4) - 7 (y + 1) + 7 (z - 2) = 0$

$\Rightarrow (x - 4) + (y + 1) - (z - 2) = 0$

$\Rightarrow x + y - z - 1 = 0$

$(1, 1, 1)$ is the only point that satisfies.

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3D Geometry 3 Question 24

24. The plane passing through the point $(4, - 1, 2)$ and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point

Answer:

Solution:

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3D Geometry 3 Question 24

24. The plane passing through the point (4,−1,2) and parallel to the lines x+23=y−2−1=z+12 and x−21=y−32=z−43 also passes through the point

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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24. The plane passing through the point $(4, - 1, 2)$ and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point