Three Dimensional Geometry jee mains questions with solutions

Q 91 :

Let a line $l$ pass through the origin and be perpendicular to the lines, $l_{1} : \vec{r} = (\hat{i} - 11 \hat{j} - 7 \hat{k}) + λ (\hat{i} + 2 \hat{j} + 3 \hat{k}), λ \in ℝ$ and $l_{2} : \vec{r} = (- \hat{i} + \hat{k}) + μ (2 \hat{i} + 2 \hat{j} + \hat{k}), μ \in ℝ .$ If P is the point of intersection of $l$ and $l_{1}$ , and $Q (α, β, γ)$ is the foot of the perpendicular from $P$ on $l_{2}$ , then $9 (α + β + γ)$ is equal to _______ . [2023]

0%

(5)

$Let l = (0 \hat{i} + 0 \hat{j} + 0 \hat{k}) + γ (a \hat{i} + b \hat{j} + c \hat{k}) = γ (a \hat{i} + b \hat{j} + c \hat{k})$

Since $l$ is perpendicular to the lines $l_{1}$ and $l_{2}$ ,

$a \hat{i} + b \hat{j} + c \hat{k} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 2 & 1 \end{matrix} |$ $= \hat{i} (2 - 6) - \hat{j} (1 - 6) + \hat{k} (2 - 4)$

$= - 4 \hat{i} + 5 \hat{j} - 2 \hat{k}$ $∴$ $l = γ (- 4 \hat{i} + 5 \hat{j} - 2 \hat{k})$

$P$ is the point of intersection of $l$ and $l_{1}$ .

$- 4 γ = 1 + λ, 5 γ = - 11 + 2 λ, - 2 γ = - 7 + 3 λ$

$By solving these equations, we get γ = - 1 .$

$So, P \equiv (4, - 5, 2)$

$Let Q (- 1 + 2 μ, 2 μ, 1 + μ)$

$Then, \vec{P Q} \cdot (2 \hat{i} + 2 \hat{j} + \hat{k}) = 0$

$\Rightarrow [(2 μ - 5) \hat{i} + (2 μ + 5) \hat{j} + (μ - 1) \hat{k}] \cdot (2 \hat{i} + 2 \hat{j} + \hat{k}) = 0$

$\Rightarrow 4 μ - 10 + 4 μ + 10 + μ - 1 = 0 \Rightarrow 9 μ = 1 \Rightarrow μ = \frac{1}{9}$

$Hence, 9 (α + β + γ) = 9 (\frac{- 7}{9} + \frac{2}{9} + \frac{10}{9}) = 5$

Q 92 :

Let the plane P contain the line $2 x + y - z - 3 = 0 = 5 x - 3 y + 4 z + 9$ and be parallel to the line $\frac{x + 2}{2} = \frac{3 - y}{- 4} = \frac{z - 7}{5} .$ Then the distance of the point $A (8, - 1, - 19)$ from the plane P measured parallel to the line $\frac{x}{- 3} = \frac{y - 5}{4} = \frac{2 - z}{- 12}$ is equal to _______ . [2023]

0%

(26)

$Equation of the plane passing through the line of intersection of the given planes is$

$(2 x + y - z - 3) + λ (5 x - 3 y + 4 z + 9) = 0$

$\Rightarrow x (2 + 5 λ) + y (1 - 3 λ) + z (4 λ - 1) + 9 λ - 3 = 0 \dots (i)$

$The plane is parallel to the line$

$\frac{x + 2}{2} = \frac{3 - y}{- 4} = \frac{z - 7}{5}$ $i.e.,$ $\frac{x + 2}{2} = \frac{y - 3}{4} = \frac{z - 7}{5}$

$∴$ $2 (2 + 5 λ) + 4 (1 - 3 λ) + 5 (4 λ - 1) = 0$

$\Rightarrow 18 λ + 3 = 0 \Rightarrow λ = - \frac{1}{6}$

$Putting the value λ = - \frac{1}{6} in equation (i), we get$

$x (2 - \frac{5}{6}) + y (1 + \frac{3}{6}) + z (\frac{- 4}{6} - 1) - \frac{9}{6} - 3 = 0$

$\Rightarrow 7 x + 9 y - 10 z - 27 = 0$

$This is the required equation of plane.$

$Now, the given point is (8, - 1, - 19) and the equation of parallel line is \frac{x}{- 3} = \frac{y - 5}{4} = \frac{z - 2}{12}$

$∴$ $Equation of line passing through the given point and parallel to the line is$

$\frac{x - 8}{- 3} = \frac{y + 1}{4} = \frac{z + 19}{12} = k$

$\Rightarrow x = - 3 k + 8, y = 4 k - 1, z = 12 k - 19$

$This will satisfy the plane: 7 x + 9 y - 10 z - 27 = 0$

$- 21 k + 56 + 36 k - 9 - 120 k + 190 - 27 = 0$

$\Rightarrow - 105 k + 210 = 0 \Rightarrow k = 2$

$∴$ $x = 2, y = 7, z = 5$

$∴$ $Required distance = \sqrt{{(8 - 2)}^{2} + {(- 1 - 7)}^{2} + {(- 19 - 5)}^{2}}$

$= \sqrt{36 + 64 + 576} = \sqrt{676} = 26 units$

Q 93 :

Let a line L pass through the point P(2, 3, 1) and be parallel to the line $x + 3 y - 2 z - 2 = 0 = x - y + 2 z .$ If the distance of L from the point (5, 3, 8) is $α$ , then $3 α^{2}$ is equal to _________ . [2023]

0%

(158)

$Let \vec{a} = \hat{i} + 3 \hat{j} - 2 \hat{k} and \vec{b} = \hat{i} - \hat{j} + 2 \hat{k}$

$\vec{a} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & - 2 \\ 1 & - 1 & 2 \end{matrix} |$ $= 4 \hat{i} - 4 \hat{j} - 4 \hat{k} = 4 (\hat{i} - \hat{j} - \hat{k})$

$Line will be parallel to \vec{a} \times \vec{b} \Rightarrow \vec{n} = \hat{i} - \hat{j} - \hat{k}$

$Equation of the line passing through the point P (2, 3, 1) and parallel to \vec{a} \times \vec{b}$

$∴$ $\frac{x - 2}{1} = \frac{y - 3}{- 1} = \frac{z - 1}{- 1}$

${\vec{a}}_{1} = 2 \hat{i} + 3 \hat{j} + \hat{k}, {\vec{a}}_{2} = 5 \hat{i} + 3 \hat{j} + 8 \hat{k}$ $\Rightarrow {\vec{a}}_{2} - {\vec{a}}_{1} = 3 \hat{i} + 7 \hat{k}$

$({\vec{a}}_{2} - {\vec{a}}_{1}) \times \vec{n} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & 7 \\ 1 & - 1 & - 1 \end{matrix} |$ $= 7 \hat{i} - \hat{j} (- 3 - 7) + \hat{k} (- 3)$

$= 7 \hat{i} + 10 \hat{j} - 3 \hat{k}$

$Now, α =$ $| \frac{({\vec{a}}_{2} - {\vec{a}}_{1}) \times \vec{n}}{| \vec{n} |} |$ $= \frac{| 7 \hat{i} + 10 \hat{j} - 3 \hat{k} |}{\sqrt{1 + 1 + 1}}$

$= \frac{\sqrt{49 + 100 + 9}}{\sqrt{3}} = \frac{\sqrt{158}}{\sqrt{3}} \Rightarrow 3 α^{2} = 158 units$

Q 94 :

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1 - y}{- 2} = \frac{z - 2}{3} .$ The coordinates of A and B are (1,6,3) and (4,9, $α$ ) respectively and C is at a distance of 10 units from B. The area (in sq. units) of $∆$ ABC is: [2026]

$5 \sqrt{13}$
$10 \sqrt{13}$
$20 \sqrt{13}$
$15 \sqrt{13}$

1

2

3

4

(1)

$\frac{4}{1} = \frac{9 - 1}{2} = \frac{α - 2}{3} \Rightarrow α = 14$

$\vec{A D} \cdot (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 0$

$(λ - 1) \hat{i} + (2 λ - 5) \hat{j} + (3 λ - 1) \hat{k} = \vec{A D}$

$\Rightarrow λ - 1 + 4 λ - 10 + 9 λ - 3 = 0$

$\Rightarrow 14 λ = 14 \Rightarrow λ = 1$

$D = (1, 3, 5)$

$A D = \sqrt{3^{2} + 2^{2}} = \sqrt{13}$

$Ar (∆ A B C) = \frac{1}{2} \times \sqrt{13} \times 10 = 5 \sqrt{13}$

Q 95 :

Let the lines $L_{1} : \vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k}), λ \in ℝ$ and $L_{2} : \vec{r} = (4 \hat{i} + \hat{j}) + μ (5 \hat{i} + 2 \hat{j} + \hat{k}), μ \in ℝ,$ intersect at the point R. Let P and Q be the points lying on lines $L_{1}$ and $L_{2}$ respectively, such that $| \vec{P R} |$ $= \sqrt{29}$ and $| \vec{P Q} |$ $= \sqrt{\frac{47}{3}}$ . If the point P lies in the first octant, then $27 {(Q R)}^{2}$ is equal to [2026]

320
340
360
348

1

2

3

4

(3)

$For POI$

$2 λ + 1 = 5 μ + 4; 3 λ + 2 = 2 μ + 1; 4 λ + 3 = μ$

$\Rightarrow λ = μ = - 1$

$R (- 1, - 1, - 1), P (2 λ + 1, 3 λ + 2, 4 λ + 3)$

$P R^{2} = 29 \Rightarrow {(2 λ + 2)}^{2} + {(3 λ + 3)}^{2} + {(4 λ + 4)}^{2} = 29$

$\Rightarrow λ = 0 or λ = - 2 (Reject)$

$\Rightarrow P (1, 2, 3)$

$Q (5 μ + 4, 2 μ + 1, μ)$

$| P Q |$ $= \sqrt{\frac{47}{3}} \Rightarrow P Q^{2} = \frac{47}{3}$

$\Rightarrow {(5 μ + 3)}^{2} + {(2 μ - 1)}^{2} + {(μ - 3)}^{2} = \frac{47}{3}$

$\Rightarrow μ = - \frac{1}{3}$

$Q = (\frac{7}{3}, \frac{1}{3}, - \frac{1}{3})$

${(Q R)}^{2} = {(\frac{7}{3} + 1)}^{2} + {(\frac{1}{3} + 1)}^{2} + {(- \frac{1}{3} + 1)}^{2}$

$= \frac{100 + 16 + 4}{9} = \frac{120}{9}$

$\Rightarrow 27 \times {(Q R)}^{2} = 27 \times \frac{120}{9} = 360$

Q 96 :

Let a line L passing through the point P(1,1,1) be perpendicular to the lines $\frac{x - 4}{4} = \frac{y - 1}{1} = \frac{z - 1}{1} and \frac{x - 17}{1} = \frac{y - 71}{1} = \frac{z}{0} .$ Let the line L intersect the $y z$ -plane at the point Q. Another line parallel to L and passing through the point S(1, 0, -1) intersects the $y z$ -plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ________ . [2026]

0%

(6)

$d_{1} = ⟨ 4, 1, 1 ⟩$ and $d_{2} = ⟨ 1, 1, 0 ⟩$

$d_{L} = d_{1} \times d_{2} =$ $| \begin{matrix} i & j & k \\ 4 & 1 & 1 \\ 1 & 1 & 0 \end{matrix} |$ $= ⟨ - 1, 1, 3 ⟩$

$Line L passes through P ⟨ 1, 1, 1 ⟩ with direction d_{L} = ⟨ - 1, 1, 3 ⟩$

$r (t) = ⟨ 1, 1, 1 ⟩ + t ⟨ - 1, 1, 3 ⟩ = ⟨ 1 - t, 1 + t, 1 + 3 t ⟩$

$For point Q, x = 0 \Rightarrow t = 1$

$Q = ⟨ 0, 2, 4 ⟩$

$Another line parallel to L passes through S ⟨ 1, 0, - 1 ⟩$

$r' (μ) = ⟨ 1, 0, - 1 ⟩ + μ ⟨ - 1, 1, 3 ⟩ = ⟨ 1 - μ, μ, - 1 + 3 μ ⟩$

$For point R, x = 0 \Rightarrow μ = 1$

$R = ⟨ 0, 1, 2 ⟩$

$Area of parallelogram with adjacent vectors \vec{P Q} and \vec{P S}$

$\vec{P Q} = ⟨ - 1, 1, 3 ⟩$

$\vec{P S} = ⟨ 0, - 1, - 2 ⟩$

Area of parallelogram

$\vec{P Q} \times \vec{P S} =$ $| \begin{matrix} i & j & k \\ - 1 & 1 & 3 \\ 0 & - 1 & - 2 \end{matrix} |$ $= ⟨ 1, - 2, 1 ⟩$

$Area = \sqrt{1^{2} + {(- 2)}^{2} + 1^{2}} = \sqrt{6}$

Q 97 :

If the distance of the point $P (43, α, β), β < 0,$ from the line $\vec{r} = 4 \hat{i} - \hat{k} + μ (2 \hat{i} + 3 \hat{k}), μ \in ℝ$ along a line with direction ratios $3, - 1, 0$ is $13 \sqrt{10},$ then $α^{2} + β^{2}$ is equal to _______ [2026]

0%

(170)

$\frac{x - 43}{3} = \frac{y - α}{- 1} = \frac{z - β}{0} \Rightarrow P_{1} (43 + 3 λ, α - λ, β)$

$\frac{x - 4}{2} = \frac{y}{0} = \frac{z + 1}{3} \Rightarrow P_{1} (2 μ + 4, 0, 3 μ - 1)$

$∴$ $μ = \frac{3 λ + 39}{2}, α = λ, β = \frac{9 λ - 115}{2}$

$P (43, α, β), P_{1} (43 + 3 α, 0, β)$

${(P P_{1})}^{2} = 1690 = 10 α^{2} \Rightarrow α = 13, β = 1$

$∴$ $α^{2} + β^{2} = 170$

Q 98 :

The sum of all values of $α$ , for which the shortest distance between the lines

$\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α} and \frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $\sqrt{2}$ , is [2026]

8
- 6
6
- 8

1

2

3

4

(2)

$\sqrt{2} =$ $\frac{| \begin{matrix} - 1 & 1 & 3 \\ α & - 1 & - α \\ α & 2 & 2 α \end{matrix} |}{| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ α & - 1 & - α \\ α & 2 & 2 α \end{matrix} |}$

$\sqrt{2} = \frac{- 1 (- 2 α + 2 α) - 1 (2 α^{2} + α^{2}) + 3 (2 α + α)}{| \hat{i} (- 2 α + 2 α) - \hat{j} (2 α^{2} + α^{2}) + \hat{k} (2 α + α) |}$

$\sqrt{2} = \frac{- 3 α^{2} + 9 α}{\sqrt{9 α^{4} + 9 α^{2}}}$

$\sqrt{2} = \frac{- α + 3}{\sqrt{α^{2} + 1}}$

$\Rightarrow 2 α^{2} + 2 = α^{2} + 9 - 6 α$

$α^{2} + 6 α - 7 = 0$

$(α + 7) (α - 1) = 0$

$α = - 7, 1$

$sum = - 7 + 1 = - 6$

option (2)

Q 99 :

If the image of the point $P (a, 2, a)$ in the line $\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1}$ is Q and the image of Q in the line $\frac{x - 2 b}{2} = \frac{y - a}{1} = \frac{z + 2 b}{- 5}$ is P, then $a + b$ is equal to ______. [2026]

0%

(3)

$\frac{2 λ - a + a}{2} = - 5 μ - 2 b$

$and \frac{a + 4 λ - a}{2} = 2 μ + 2 b$

$and \frac{2 λ - 2 a - 2 + 2}{2} = μ + a$

$\Rightarrow λ - μ = b$

$λ - μ = 2 a$

$λ + 5 μ = - 2 b$

$\Rightarrow b = 2 a and λ = a, μ = - a$

$(a \hat{i} - 2 \hat{j} + 0 \hat{k}) \cdot (2 \hat{i} + \hat{j} + \hat{k}) = 0$

$2 a - 2 = 0$

$a = 1$

$b = 2 a = 2 \times 1 = 2$

$a + b = 1 + 2 = 3$

Q 100 :

Let the line $L_{1}$ be parallel to the vector $- 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and pass through the point (2,6,7), and the line $L_{2}$ be parallel to the vector $2 \hat{i} + \hat{j} + 3 \hat{k}$ and pass through the point (4,3,5). If the line $L_{3}$ is parallel to the vector $- 3 \hat{i} + 5 \hat{j} + 16 \hat{k}$ and intersects the lines $L_{1} and L_{2}$ at the points C and D, respectively, then $| \vec{C D} |^{2}$ is equal to: [2026]

171
89
290
312

1

2

3

4

(3)

$L_{1} : \frac{x - 2}{- 3} = \frac{y - 6}{2} = \frac{z - 7}{4}$

Point $C$ on $L_{1}$ : $(- 3 λ_{1} + 2, 2 λ_{1} + 6, 4 λ_{1} + 7)$

$L_{2} : \frac{x - 4}{2} = \frac{y - 3}{1} = \frac{z - 5}{3}$

Point $D$ on $L_{2}$ : $(2 λ_{2} + 4, λ_{2} + 3, 3 λ_{2} + 5)$

D.R's of line $L_{3}$ :

$L_{3} : \frac{2 λ_{2} + 3 λ_{1} + 2}{- 3} = \frac{λ_{2} - 2 λ_{1} - 3}{5} = \frac{3 λ_{2} - 4 λ_{1} - 2}{16}$

$λ_{1} = - 3, λ_{2} = 2$

$C (11, 0, - 5)$

$D (8, 5, 11)$

$| \vec{C D} |^{2} = 3^{2} + 5^{2} + 16^{2} = 290$

Topic Question Set

0%

0%

Q 93 :

Let a line L pass through the point P(2, 3, 1) and be parallel to the line $x + 3 y - 2 z - 2 = 0 = x - y + 2 z .$ If the distance of L from the point (5, 3, 8) is $α$ , then $3 α^{2}$ is equal to _________ . [2023]

0%

Q 94 :

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1 - y}{- 2} = \frac{z - 2}{3} .$ The coordinates of A and B are (1,6,3) and (4,9, $α$ ) respectively and C is at a distance of 10 units from B. The area (in sq. units) of $∆$ ABC is: [2026]

0%

Q 97 :

If the distance of the point $P (43, α, β), β < 0,$ from the line $\vec{r} = 4 \hat{i} - \hat{k} + μ (2 \hat{i} + 3 \hat{k}), μ \in ℝ$ along a line with direction ratios $3, - 1, 0$ is $13 \sqrt{10},$ then $α^{2} + β^{2}$ is equal to _______ [2026]

0%

Q 98 :

The sum of all values of $α$ , for which the shortest distance between the lines

$\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α} and \frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $\sqrt{2}$ , is [2026]

Q 99 :

If the image of the point $P (a, 2, a)$ in the line $\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1}$ is Q and the image of Q in the line $\frac{x - 2 b}{2} = \frac{y - a}{1} = \frac{z + 2 b}{- 5}$ is P, then $a + b$ is equal to ______. [2026]

0%

Want Us to Email you About Special Offers & Updates?

Topic Question Set

0%

Q 92 : Let the plane P contain the line 2x+y-z-3=0=5x-3y+4z+9 and be parallel to the line x+22=3-y-4=z-75. Then the distance of the point A(8,-1,-19) from the plane P measured parallel to the line x-3=y-54=2-z-12 is equal to _______ . [2023]

0%

Q 93 : Let a line L pass through the point P(2, 3, 1) and be parallel to the line x+3y-2z-2=0=x-y+2z. If the distance of L from the point (5, 3, 8) is α, then 3α2 is equal to _________ . [2023]

0%

Q 94 : The vertices B and C of a triangle ABC lie on the line x1=1-y-2=z-23. The coordinates of A and B are (1,6,3) and (4,9,α) respectively and C is at a distance of 10 units from B. The area (in sq. units) of ∆ABC is: [2026]

0%

Q 97 : If the distance of the point P(43,α,β), β<0, from the line r→=4i^-k^+μ(2i^+3k^), μ∈ℝ along a line with direction ratios 3, -1, 0 is 1310, then α2+β2 is equal to _______ [2026]

0%

Q 98 : The sum of all values of α, for which the shortest distance between the lines x+1α=y-2-1=z-4-α and xα=y-12=z-12α is 2, is [2026]

Q 99 : If the image of the point P(a,2,a) in the line x2=y+a1=z1 is Q and the image of Q in the line x-2b2=y-a1=z+2b-5 is P, then a+b is equal to ______. [2026]

0%

Want Us to Email you About Special Offers & Updates?

Q 93 :

Let a line L pass through the point P(2, 3, 1) and be parallel to the line $x + 3 y - 2 z - 2 = 0 = x - y + 2 z .$ If the distance of L from the point (5, 3, 8) is $α$ , then $3 α^{2}$ is equal to _________ . [2023]

Q 94 :

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1} = \frac{1 - y}{- 2} = \frac{z - 2}{3} .$ The coordinates of A and B are (1,6,3) and (4,9, $α$ ) respectively and C is at a distance of 10 units from B. The area (in sq. units) of $∆$ ABC is: [2026]

Q 97 :

If the distance of the point $P (43, α, β), β < 0,$ from the line $\vec{r} = 4 \hat{i} - \hat{k} + μ (2 \hat{i} + 3 \hat{k}), μ \in ℝ$ along a line with direction ratios $3, - 1, 0$ is $13 \sqrt{10},$ then $α^{2} + β^{2}$ is equal to _______ [2026]

Q 98 :

The sum of all values of $α$ , for which the shortest distance between the lines

$\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α} and \frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $\sqrt{2}$ , is [2026]

Q 99 :

If the image of the point $P (a, 2, a)$ in the line $\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1}$ is Q and the image of Q in the line $\frac{x - 2 b}{2} = \frac{y - a}{1} = \frac{z + 2 b}{- 5}$ is P, then $a + b$ is equal to ______. [2026]