Three Dimensional Geometry jee mains questions with solutions

Q 81 :

The foot of perpendicular of the point (2, 0, 5) on the line $\frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{- 1}$ is $(α, β, γ)$ . Then, which of the following is not correct? [2023]

$\frac{γ}{α} = \frac{5}{8}$
$\frac{β}{γ} = - 5$
$\frac{α β}{γ} = \frac{4}{15}$
$\frac{α}{β} = - 8$

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(2)

$D.r's of line A B < α - 2, β, γ - 5 >$

Since $A B$ is perpendicular to

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

$∴$ $2 (α - 2) + 5 β - 1 (γ - 5) = 0$

$\Rightarrow 2 α + 5 β - γ + 1 = 0$ ...(i)

Also, $α, β, γ$ lies on the given line.

$∴$ $\frac{α + 1}{2} = \frac{β - 1}{5} = \frac{γ + 1}{- 1}$ ...(ii)

From (ii), (taking the first two and last two terms)

$α = \frac{2 β - 7}{5}, γ = \frac{- β - 4}{5}$

Putting these in (i), we get $4 β - 14 + 25 β + β + 4 + 5 = 0$

$\Rightarrow β = \frac{1}{6}$

which gives $α = \frac{- 4}{3}$ and $γ = \frac{- 5}{6}$ $\Rightarrow \frac{γ}{α} = \frac{\frac{- 5}{6}}{\frac{- 4}{3}} = \frac{5}{8}$ ,

$\frac{β}{γ} = \frac{1}{6} \times \frac{- 6}{5} = \frac{- 1}{5} \Rightarrow \frac{α β}{γ} = \frac{4}{15}, \frac{α}{β} = - 8, \frac{β}{γ} \neq - 5$

Q 82 :

If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersect at the point P, then the distance of the point P from the plane $z = a$ is [2023]

10
28
22
16

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(2)

$Let two lines L_{1} and L_{2} intersect at point P .$ Then we have the points on line $L_{1}$ are as $= (λ + 1, 2 λ + 2, λ - 3)$ and we have the points on line $L_{2}$ are as $= (2 μ + a, 3 μ - 2, μ + 3)$

${∵ L_{1} = \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1} = λ (say) and L_{2} = \frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1} = μ (say)}$

$Now if L_{1} and L_{2} intersect, then we must have$

$λ - 3 = μ + 3 \Rightarrow λ = μ + 6 ...(i)$

and $2 λ + 2 = 3 μ - 2 \Rightarrow 2 λ = 3 μ - 4 ...(ii)$

On solving (i) and (ii), we get $μ = 16$ and $λ = 22$

So, the coordinates of $P$ must be $(23, 46, 19)$ $\Rightarrow a = - 9$

Hence, the distance of the point $P$ from $z = - 9$ is $28 .$

Q 83 :

The shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is [2023]

$3 \sqrt{3}$
$2 \sqrt{3}$
$5 \sqrt{3}$
$4 \sqrt{3}$

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(4)

$Since two lines L_{1} and L_{2} are given as$

$L_{1} \equiv \frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}, L_{2} \equiv \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$

$For line L_{1}, we have \vec{a} = \hat{i} - 8 \hat{j} + 4 \hat{k}$

$For line L_{2}, we have \vec{b} = \hat{i} + 2 \hat{j} + 6 \hat{k}$

$\vec{p} = 2 \hat{i} - 7 \hat{j} + 5 \hat{k}$ and $\vec{q} = 2 \hat{i} + \hat{j} - 3 \hat{k}$

$So, we have$ $\vec{p} \times \vec{q} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 7 & 5 \\ 2 & 1 & - 3 \end{matrix} |$

$= \hat{i} (21 - 5) - \hat{j} (- 6 - 10) + \hat{k} (2 + 14)$

$= 16 \hat{i} + 16 \hat{j} + 16 \hat{k} = 16 (\hat{i} + \hat{j} + \hat{k})$

$The shortest distance, d$

$=$ $| \frac{(\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q})}{| \vec{p} \times \vec{q} |} |$

$=$ $| \frac{[\hat{i} (1 - 1) + \hat{j} (- 8 - 2) + \hat{k} (4 - 6)] \cdot 16 (\hat{i} + \hat{j} + \hat{k})}{\sqrt{16^{2} + 16^{2} + 16^{2}}} |$

$=$ $| \frac{(- 10 \hat{j} - 2 \hat{k}) \cdot (16 \hat{i} + 16 \hat{j} + 16 \hat{k})}{16 \sqrt{3}} |$

$=$ $| \frac{(- 10) (16) + (- 2) (16)}{16 \sqrt{3}} |$ $=$ $| \frac{192}{16 \sqrt{3}} |$ $=$ $| \frac{- 12}{\sqrt{3}} |$ $= 4 \sqrt{3}$

Q 84 :

Let the shortest distance between the lines $L : \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1}, λ \geq 0$ and $L_{1} : x + 1 = y - 1 = 4 - z$ be $2 \sqrt{6}$ . If $(α, β, γ)$ lies on L, then which of the following is NOT possible? [2023]

$α + 2 γ = 24$
$2 α + γ = 7$
$α - 2 γ = 19$
$2 α - γ = 9$

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(1)

$Since shortest distance between lines L and L_{1} is 2 \sqrt{6} .$

$Here, L : \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1}, λ \geq 0$

$and L_{1} : \frac{x + 1}{1} = \frac{y - 1}{1} = \frac{z - 4}{- 1}$

$Shortest distance between L and L_{1} is given by$

$d =$ $| \frac{| \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} |}{\sqrt{{(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2} + {(a_{1} b_{2} - a_{2} b_{1})}^{2}}} |$

$∴$ $2 \sqrt{6} =$ $| \frac{| \begin{matrix} - 1 - 5 & 1 - λ & 4 + λ \\ - 2 & 0 & 1 \\ 1 & 1 & - 1 \end{matrix} |}{\sqrt{{(0 - 1)}^{2} + {(1 - 2)}^{2} + {(2 - 0)}^{2}}} |$

$=$ $| \frac{- 6 (- 1) - (- λ + 1) (2 - 1) + (λ + 4) (- 2)}{\sqrt{1 + 1 + 4}} |$

$=$ $| \frac{6 + λ - 1 - 2 λ - 8}{\sqrt{6}} |$ $=$ $| \frac{- λ - 3}{\sqrt{6}} |$

$\Rightarrow \frac{- 3 - λ}{\sqrt{6}} = \pm 2 \sqrt{6} \Rightarrow \frac{- 3 - λ}{\sqrt{6}} = 2 \sqrt{6} and \frac{- 3 - λ}{\sqrt{6}} = - 2 \sqrt{6}$

$- 3 - λ = 12 \Rightarrow λ = - 15$ and $- 3 - λ = - 12 \Rightarrow λ = 9$

$Since λ \geq 0, therefore λ = 9$

$Now, from line L, \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1} = r (say)$

$\Rightarrow x = - 2 r + 5, y = 0 + λ, z = r - λ$

$\Rightarrow x = - 2 r + 5, y = 9, z = r - 9$

$Since (α, β, γ) lies on L,$ then $α = - 2 r + 5, β = 9, γ = r - 9$

$Now, α + 2 γ = - 2 r + 5 + 2 r - 18 = - 13$

$2 α + γ = - 4 r + 10 + r - 9 = - 3 r + 1, r \in R$

$α - 2 γ = - 2 r + 5 - 2 r + 18 = - 4 r + 23, r \in R$

$2 α - γ = - 4 r + 10 - r + 9 = - 5 r + 19, r \in R$

$Hence, α + 2 γ = 24 is not possible.$

Q 85 :

If the lines $\frac{x - 1}{2} = \frac{2 - y}{- 3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect, then the magnitude of the minimum value of $8 α β$ is _______ . [2023]

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(18)

$Let \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{α} = λ (say)$ ...(i)

$and \frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β} = μ (say)$ ...(ii)

$∴$ $Any point on line (i) and (ii) are of the forms (2 λ + 1, 3 λ + 2, α λ + 3)$ $and (5 μ + 4, 2 μ + 1, β μ) respectively .$

$The lines are intersecting.$

$∴$ $2 λ + 1 = 5 μ + 4, 3 λ + 2 = 2 μ + 1, α λ + 3 = β μ$

$Solving first two equations, we get λ = - 1 and μ = - 1$

$From third equation, we have$

$- α + 3 = - β \Rightarrow α - 3 = β$ ...(iii)

$Let y = 8 α β = 8 α (α - 3)$

$y' = 8 α + 8 (α - 3) = 16 α - 24$

$For maxima/minima, 16 α - 24 = 0 \Rightarrow α = \frac{3}{2}$

$Now, y'' = 16 > 0$

$∴$ $8 α β is minimum at α = \frac{3}{2}$

$So, minimum value of$ $| 8 α β |$ $=$ $| 8 \times \frac{3}{2} \times \frac{- 3}{2} |$ $= 18$

Q 86 :

If the line $x = y = z$ intersects the line $x \sin A + y \sin B + z \sin C - 18 = 0 = x \sin 2 A + y \sin 2 B + z \sin 2 C - 9,$ where A, B, C are the angles of a triangle ABC, then $80 (\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2})$ is equal to _________ . [2023]

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(5)

$Any point on the given line \frac{x}{1} = \frac{y}{1} = \frac{z}{1} = λ is (λ, λ, λ)$

$If it intersects the given lines then it must satisfy them.$

$\Rightarrow λ (\sin A + \sin B + \sin C) = 2 \times 3^{2}$ ...(i)

$and λ (\sin 2 A + \sin 2 B + \sin 2 C) = 3^{2}$ ...(ii)

$On dividing equation (ii) by (i), we get$

$\frac{\sin 2 A + \sin 2 B + \sin 2 C}{\sin A + \sin B + \sin C} = \frac{1}{2}$ $\Rightarrow \frac{4 \sin A \sin B \sin C}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} = \frac{1}{2}$

$\Rightarrow 8 [\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}] = \frac{1}{2} \Rightarrow \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16}$

$∴$ $80 (\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}) = 80 \times \frac{1}{16} = 5$

Q 87 :

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______ . [2023]

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(14)

$L_{1} : \frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$

$L_{2} : \frac{x - 6}{3} = \frac{y - 1}{- 2} = \frac{z + 8}{0}$

$Shortest distance =$ $\frac{| | \begin{matrix} 6 - 2 & 1 - (- 1) & - 8 - 6 \\ 3 & 2 & 2 \\ 3 & - 2 & 0 \end{matrix} | |}{| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & - 2 & 0 \end{matrix} | |}$

$=$ $\frac{| | \begin{matrix} 4 & 2 & - 14 \\ 3 & 2 & 2 \\ 3 & - 2 & 0 \end{matrix} | |}{| 4 \hat{i} + 6 \hat{j} - 12 \hat{k} |}$ $= \frac{4 (0 + 4) - 2 (0 - 6) - 14 (- 6 - 6)}{\sqrt{4^{2} + 6^{2} + 12^{2}}}$

$= \frac{16 + 12 + 14 \times 12}{14} = 14 units.$

Q 88 :

If the shortest distance between the lines $\frac{x + \sqrt{6}}{2} = \frac{y - \sqrt{6}}{3} = \frac{z - \sqrt{6}}{4}$ and $\frac{x - λ}{3} = \frac{y - 2 \sqrt{6}}{4} = \frac{z + 2 \sqrt{6}}{5}$ is $6$ , then the square of the sum of all possible values of $λ$ is ________ . [2023]

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(384)

$Given, shortest distance between the lines$

$\frac{x + \sqrt{6}}{2} = \frac{y - 2 \sqrt{6}}{3} = \frac{z - \sqrt{6}}{4}$

$and \frac{x - λ}{3} = \frac{y - 2 \sqrt{6}}{4} = \frac{z + 2 \sqrt{6}}{5} is 6 .$

$Vector along line of shortest distance$

$=$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix} |$ $= \hat{i} (15 - 16) - \hat{j} (10 - 12) + \hat{k} (8 - 9)$

$= - \hat{i} + 2 \hat{j} - \hat{k} (its magnitude is \sqrt{6})$

$Now, \frac{1}{\sqrt{6}}$ $| \begin{matrix} λ + \sqrt{6} & \sqrt{6} & - 3 \sqrt{6} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix} |$ $= \pm 6$ $\Rightarrow λ = - 2 \sqrt{6}, 10 \sqrt{6}$

$So, square of sum of these values: {(10 \sqrt{6} - 2 \sqrt{6})}^{2} = {(8 \sqrt{6})}^{2} = 384$

Q 89 :

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ , and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $α$ , then $28 α^{2}$ is equal to ______ . [2023]

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(18)

$The equation of the line passing through (1, 2, 3) and (2, 3, 4) is$

$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} + \hat{j} + \hat{k}) (∵ \vec{r} = \vec{a} + λ \vec{p})$

and vector form of equation of given second line is

$\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + μ (2 \hat{i} - \hat{j}) (∵ \vec{r} = \vec{b} + μ \vec{q})$

$Now,$ $\vec{p} \times \vec{q} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & - 1 & 0 \end{matrix} |$ $= \hat{i} + 2 \hat{j} - 3 \hat{k}$

$So, the shortest distance =$ $| \frac{(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q})}{| \vec{p} \times \vec{q} |} |$

$=$ $| \frac{(- 3 \hat{j} - \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 3 \hat{k})}{\sqrt{14}} |$ $=$ $| \frac{- 6 + 3}{\sqrt{14}} |$ $= \frac{3}{\sqrt{14}} = α = \frac{3}{\sqrt{14}}$

$Now, 28 α^{2} = 28 \times \frac{9}{14} = 18$

Q 90 :

Let the co-ordinates of one vertex of $Δ A B C$ be $A (0, 2, α)$ and the other two vertices lie on the line $\frac{x + α}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} .$ For $α \in Z$ , if the area of $Δ A B C$ is $21$ sq. units and the line segment $B C$ has length $2 \sqrt{21}$ units, then $α^{2}$ is equal to _________ . [2023]

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