Three Dimensional Geometry jee mains questions with solutions

Q 11 :

The vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ is:

$\vec{r} \cdot (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = 6$
$\vec{r} \cdot (\frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k}) = \frac{6}{\sqrt{29}}$
Both (a) and (b)
None of the above

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(2)

Ans. $\vec{r} \cdot (\frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k}) = \frac{6}{\sqrt{29}}$

Explanation:

Let $\vec{n} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} .$ Then,

$\hat{n} = \frac{\vec{n}}{| \vec{n} |} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{4 + 9 + 16}} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{29}}$

Hence, the required equation of the plane is

$\vec{r} \cdot (\frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k}) = \frac{6}{\sqrt{29}}$

Q 12 :

A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is $(α, β, γ)$ . The equation of the plane is:

$\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 3$
$α x + β y + γ y = 3 α β γ$
$\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = \frac{1}{2}$
None of these

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(1)

Ans. $\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 3$

Explanation:

Let the equation of the required plane be

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ ...(i)

It meets coordinate axes in points

$A (a, 0, 0), B (0, b, 0), C (0, 0, c)$

The centroid of $∆ A B C$ is $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$

$\Rightarrow$ $\frac{a}{3} = α, \frac{b}{3} = β, \frac{c}{3} = γ$

$\Rightarrow$ $a = 3 α, b = 3 β, c = 3 γ$

Hence, the required plane is

$\frac{x}{3 α} + \frac{y}{3 β} + \frac{z}{3 γ} = 1$

i.e., $\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 3$

Q 13 :

If vector equation of the plane $\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$ is $\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} + p \hat{k})$ , then $p$ is equal to:

0
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3

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(1)

Ans. 0

Explanation:

The given line is

$\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$

$\Rightarrow$ $\frac{x - 2}{2} = \frac{y - \frac{5}{2}}{- \frac{3}{2}} = \frac{z + 1}{0}$

This shows that the given line passes through the point $(2, \frac{5}{2}, - 1)$ and has direction ratios $(2, - \frac{3}{2}, 0) .$ Thus, given line passes through the point having position vector $\vec{a} = 2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}$ and is parallel to the vector $\vec{b} = (2 \hat{i} - \frac{3}{2} \hat{j} - 0 \hat{k}) .$

So, its vector equation is

$\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} - 0 \hat{k})$

Hence, $p = 0$

Topic Question Set

Q 11 :

The vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ is:

Q 12 :

A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is $(α, β, γ)$ . The equation of the plane is:

Q 13 :

If vector equation of the plane $\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$ is $\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} + p \hat{k})$ , then $p$ is equal to:

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Topic Question Set

Q 11 : The vector equation of the plane which is at a distance of 629 from the origin and its normal vector from the origin is 2i^-3j^+4k^ is:

Q 12 : A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is (α,β,γ). The equation of the plane is:

Q 13 : If vector equation of the plane x-22=2y-5-3=z+1, is r→=(2i^+52j^-k^)+λ(2i^-32j^+pk^), then p is equal to:

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Q 11 :

The vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ is:

Q 12 :

A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is $(α, β, γ)$ . The equation of the plane is:

Q 13 :

If vector equation of the plane $\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$ is $\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} + p \hat{k})$ , then $p$ is equal to: