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]

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(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]

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(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]

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(158)

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

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

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

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

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

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

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

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

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

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

Topic Question Set

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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]

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