Three Dimensional Geometry jee mains questions with solutions

Q 1 :

Each of the angles $β$ and $γ$ that a given line makes with the positive y-axes and z-axes, respectively, is half of the angle that this line makes with the positive x-axes. Then the sum of all possible values of the angle $β$ is [2025]

$π$
$\frac{π}{2}$
$\frac{3 π}{4}$
$\frac{3 π}{2}$

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(3)

Let the line makes angle $α$ with positive x-axis, then $β = \frac{α}{2}$ and $γ = \frac{α}{2}$

Now, $\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$

$\Rightarrow \cos^{2} α + 2 \cos^{2} \frac{α}{2} = 1$

$\Rightarrow \cos^{2} α + \cos α = 0$

$\Rightarrow \cos α (\cos α + 1) = 0$

$\Rightarrow \cos α = 0, - 1 \Rightarrow α = \frac{π}{2}, π$

Now, $β = \frac{α}{2} \Rightarrow β = \frac{π}{4}, \frac{π}{2}$

So, required sum $= \frac{π}{4} + \frac{π}{2} = \frac{3 π}{4}$ .

Q 2 :

Let A(x, y, z) be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B = (1, 4, –1) and C = (2, 0, –2). Then among the statements

(S1) : $△$ ABC is an isosceles right angled triangle, and

(S2) : the area of $△$ ABC is $\frac{9 \sqrt{2}}{2}$ .

only (S1) is true
both are false
both are true
only (S2) is true

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(1)

Given, A(x, y, z) be a point in xy-plane. Let the point P(0, 3, 2), Q(2, 0, 3) and R(0, 0, 1)

The distance of the point AP = AQ = AR

$\Rightarrow {(x - 0)}^{2} + {(y - 3)}^{2} + {(z - 2)}^{2}$

$= {(x - 2)}^{2} + {(y - 0)}^{2} + {(z - 3)}^{2}$

$= {(x - 0)}^{2} + {(y - 0)}^{2} + {(z - 1)}^{2}$

In xy-plane, z = 0

So, $x^{2} - 4 x + y^{2} + 9 + 4 = x^{2} + y^{2} + 1 \Rightarrow - 4 x + 13 = 1 \Rightarrow x = 3$

And $x^{2} + y^{2} - 6 y + 9 + 4 = x^{2} + y^{2} + 1 \Rightarrow y = 2$

So, A(3, 2, 0), B(1, 4, –1) and C(2, 0, –2).

In $△$ ABC

$A B = \sqrt{1 + 4 + 4} = 3, B C = \sqrt{1 + 16 + 1} = \sqrt{18}, C A = \sqrt{1 + 4 + 4} = 3$

So, AB = AC and $A B^{2} + A C^{2} = {(B C)}^{2}$

$∴$ $△$ ABC is an isosceles right angled triangle.

So, (S1) is true.

Also, Area of $△ A B C = \frac{1}{2} \times (Base) (Height) = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$

So, (S2) is false.

Q 3 :

One vertex of a rectangular parallelepiped is at the origin O and the lengths of its edges along $x, y$ and $z$ axes are 3, 4 and 5 units respectively. Let P be the vertex (3, 4, 5). Then the shortest distance between the diagonal OP and an edge parallel to $z$ -axis, not passing through O or P is [2023]

$\frac{12}{5}$
$\frac{12}{5 \sqrt{5}}$
$\frac{12}{\sqrt{5}}$
$12 \sqrt{5}$

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$Equation of line O P, \frac{x - 0}{3 - 0} = \frac{y - 0}{4 - 0} = \frac{z - 0}{5 - 0}$
i.e., $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$

Now, equation of edge parallel to $z$ -axis passing through $(3, 0, 5)$ and having direction ratios $< 0, 0, 1 >$ is

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

Here, $a_{1} = (0, 0, 0), a_{2} = (3, 0, 5); b_{1} = (3, 4, 5), b_{2} = (0, 0, 1)$

So, $\vec{a_{2}} - \vec{a_{1}} = 3 \hat{i} + 5 \hat{k}$

$\vec{b_{1}} \times \vec{b_{2}} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 0 & 0 & 1 \end{matrix} |$ $= \hat{i} (4 - 0) - \hat{j} (3 - 0) + \hat{k} (0 - 0) = 4 \hat{i} - 3 \hat{j}$

$∴$ $Required shortest distance =$ $| \frac{(3 \hat{i} + 5 \hat{k}) \cdot (4 \hat{i} - 3 \hat{j})}{| 4 \hat{i} - 3 \hat{j} |} |$

$= \frac{12}{\sqrt{16 + 9}} = \frac{12}{5} units .$

Q 4 :

The area of the quadrilateral ABCD with vertices A(2, 1, 1), B(1, 2, 5), C(−2, −3, 5) and D(1, −6, −7) is equal to [2023]

$8 \sqrt{38}$
$48$
$54$
$9 \sqrt{38}$

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(1)

$The area of quadrilateral A B C D is equal to \frac{1}{2}$ $| \vec{A C} \times \vec{B D} |$ .

Now, $\vec{A C} = (- 2 \hat{i} - 3 \hat{j} + 5 \hat{k}) - (2 \hat{i} + \hat{j} + \hat{k}) = - 4 \hat{i} - 4 \hat{j} + 4 \hat{k}$

and $\vec{B D} = (\hat{i} - 6 \hat{j} - 7 \hat{k}) - (\hat{i} + 2 \hat{j} + 5 \hat{k}) = - 8 \hat{j} - 12 \hat{k}$

So, $\frac{1}{2}$ $| \vec{A C} \times \vec{B D} |$ $= \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 4 & - 4 & 4 \\ 0 & - 8 & - 12 \end{matrix} | |$

$= \frac{1}{2}$ $| 80 \hat{i} - 48 \hat{j} + 32 \hat{k} |$ $= \frac{1}{2} \sqrt{9728} = 8 \sqrt{38}$

Q 5 :

The distance of the point P(4, 6, −2) from the line passing through the point (−3, 2, 3) and parallel to a line with direction ratios 3, 3, −1 is equal to: [2023]

$2 \sqrt{3}$
$\sqrt{14}$
$3$
$\sqrt{6}$

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$Let Q = - 3 \hat{i} + 2 \hat{j} + 3 \hat{k}$

$\vec{P Q} = - 7 \hat{i} - 4 \hat{j} + 5 \hat{k}, \vec{b} = 3 \hat{i} + 3 \hat{j} - \hat{k} \Rightarrow$ $| \vec{b} |$ $= \sqrt{19}$

$\vec{P Q} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 7 & - 4 & 5 \\ 3 & 3 & - 1 \end{matrix} |$ $= - 11 \hat{i} + 8 \hat{j} - 9 \hat{k}$

$Required distance =$ $| \frac{\vec{P Q} \times \vec{b}}{| \vec{b} |} |$ $= \frac{\sqrt{11^{2} + 8^{2} + 9^{2}}}{\sqrt{19}} = \sqrt{14} units$

Q 6 :

Let the plane $x + 3 y - 2 z + 6 = 0$ meet the coordinate axes at the points A, B, C. If the orthocenter of the triangle ABC is $(α, β, \frac{6}{7})$ , then $98 {(α + β)}^{2}$ is equal to _____ . [2023]

0%

(288)

Plane $x + 3 y - 2 z + 6 = 0$ meets the coordinate axes at the points A, B, C.

$For x-axis: x + 0 - 0 + 6 = 0 \Rightarrow x = - 6 i.e., (- 6, 0, 0)$

$For y-axis: 0 + 3 y - 0 + 6 = 0 \Rightarrow y = - 2 i.e., (0, - 2, 0)$

$For z-axis: 0 + 0 - 2 z + 6 = 0 \Rightarrow z = 3 i.e., (0, 0, 3)$

$\vec{A B} = 6 \hat{i} - 2 \hat{j}, \vec{B C} = 2 \hat{j} + 3 \hat{k}, \vec{A C} = 6 \hat{i} + 3 \hat{k}$

Now, $\vec{A P} \cdot \vec{B C} = 0$

$∴$ $(α + 6, β, \frac{6}{7}) \cdot (0, 2, 3) = 0$

$\Rightarrow 2 β + 3 \cdot \frac{6}{7} = 0 \Rightarrow β = \frac{- 9}{7}$

Similarly $\vec{C P} \cdot \vec{A B} = 0$

$(α, β, - \frac{15}{7}) \cdot (6, - 2, 0) = 0 \Rightarrow 6 α - 2 β = 0$

$\Rightarrow 6 α - 2 (\frac{- 9}{7}) = 0 \Rightarrow 6 α = \frac{- 18}{7} \Rightarrow α = \frac{- 3}{7}$

$∴$ $98 {(α + β)}^{2} = 98 {(\frac{- 3}{7} - \frac{9}{7})}^{2} = 98 {(\frac{- 12}{7})}^{2} = 98 \cdot \frac{144}{49} = 288$

Q 7 :

Let the direction cosines of two lines satisfy the equations $4 l + m - n = 0$ and $2 m n + 10 n l + 3 l m = 0 .$ Then the cosine of the acute angle between these lines is: [2026]

$\frac{10}{\sqrt{38}}$
$\frac{20}{3 \sqrt{38}}$
$\frac{10}{7 \sqrt{38}}$
$\frac{10}{3 \sqrt{38}}$

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(4)

Topic Question Set

Q 1 :

Each of the angles $β$ and $γ$ that a given line makes with the positive y-axes and z-axes, respectively, is half of the angle that this line makes with the positive x-axes. Then the sum of all possible values of the angle $β$ is [2025]

Q 4 :

The area of the quadrilateral ABCD with vertices A(2, 1, 1), B(1, 2, 5), C(−2, −3, 5) and D(1, −6, −7) is equal to [2023]

Q 5 :

The distance of the point P(4, 6, −2) from the line passing through the point (−3, 2, 3) and parallel to a line with direction ratios 3, 3, −1 is equal to: [2023]

Q 6 :

Let the plane $x + 3 y - 2 z + 6 = 0$ meet the coordinate axes at the points A, B, C. If the orthocenter of the triangle ABC is $(α, β, \frac{6}{7})$ , then $98 {(α + β)}^{2}$ is equal to _____ . [2023]

0%

Q 7 :

Let the direction cosines of two lines satisfy the equations $4 l + m - n = 0$ and $2 m n + 10 n l + 3 l m = 0 .$ Then the cosine of the acute angle between these lines is: [2026]

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