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Three Dimensional Geometry

121379 Let the image of the point $P (1, 2, 6)$ in the plane passing through the points $A (1, 2, 0), B (1, 4, 1)$ and $C (0, 5, 1)$ be $Q (α^{2} + β^{2} + γ^{2})$ is equal to :

1 65

2 70

3 76

4 62

Explanation:

A Given,
$P (1, 2, 6) & A (1, 2, 0), B (1, 4, 1), C (0, 5, 1)$
Equation of required plane will be
$a (x - 1) + b (y - 2) + c (z - 0) = 0$
Putting $(1, 4, 1) \Rightarrow 2 b + c = 0$
putting $(0, 5, 1) \Rightarrow - a + 3 b + c = 0$
On solving (i) \& (ii) we get -
$a = b and c = - 2 b$
So,
$1 (x - 1) + 1 (y - 2) - 2 (z - 0) = 0$
$\Rightarrow x + y - 2 z - 3 = 0$
Image of $P$ on this plane is -
$\frac{x - 1}{1} = \frac{β - 2}{1} = \frac{γ - 6}{- 2} = \frac{- 2 (1 + 2 - 12 - 3)}{6}$
$\Rightarrow \frac{α - 1}{1} = \frac{β - 2}{1} = \frac{γ - 6}{- 2} = 4$
$\Rightarrow α = 5, β = 6, γ = - 2$
$So, α^{2} + β^{2} + γ^{2} = 25 + 36 + 4$
$\Rightarrow α^{2} + β^{2} + γ^{2} = 65$

JEE Main-10.04.2023

Three Dimensional Geometry

121381 The perpendicular distance from origin to the plane $x + 2 y - 2 z + 5 = 0$ equals units.

1

\frac{3}{5}

2

\frac{5}{3}

3

\frac{5}{9}

4 5

Explanation:

B
Distance from $(0, 0)$ to the plane $x + 2 y - 2 z + 5 = 0$ is
$d = | \frac{a x_{1} + b y_{1} + c z_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} |$
$= | \frac{1 (0) + 2 (0) - 2 (0) + 5}{\sqrt{1^{2} + 2^{2} + (- 2)^{2}}} |$
$\Rightarrow d = \frac{5}{3}$

APEAPCET-20.08.2021

Three Dimensional Geometry

121382 If $(3, 4, - 7)$ is the foot of the perpendicular drawn from the point $(- 2, 3, 6)$ to the plane $π$ then the sum of the intercepts made by the plane $π$ on the $x$ and $y$ -axes is

1 132

2 142

3 210

4 175

Explanation:

A Given,
$A (3, 4, - 7), B (- 2, 3, 6)$
$\Rightarrow \vec{AB} = - 5 \hat{i} - \hat{j} + 13 \hat{k}$
Equation of plane $λ$ containing point $A$ having normal
vector $\vec{AB}$ is
$λ : - 5 (x - 3) - 1 (y - 4) + 13 (z + 7) = 0$
$\Rightarrow - 5 x - y + 13 z = - 110$
Or $5 x + y - 13 z = 110$
$\frac{x}{22} + \frac{y}{110} - \frac{z}{110 / 13} = 1$
Sum of intercept on $x$ and $y$ axes is
$22 + 110 = 132$

AP EAMCET-06.07.2022

Three Dimensional Geometry

121383 If the point $(a, 8, - 2)$ divides the line segment joining the points $(1, 4, 6)$ and $(5, 2, 10)$ in the ratio $m$ : $n$ then $\frac{2 m}{n} - \frac{a}{3} =$

1 -7

2 1

3 -2

4 3

Explanation:

B Given, points
$A (a, 8, - 2), P (1, 4, 6)$ and $Q (5, 2, 10)$ where, ' $A$ ' divides $PQ$ is $k : 1$ ratio.
Then,
Appling section formula -
$A = (\frac{k \times 5 + 1 \times 1}{k + 1}, \frac{k \times 2 + 1 \times 4}{k + 1}, \frac{k \times 10 + 1 \times 6}{k + 1})$
$\Rightarrow (a, 8, - 2) = (\frac{5 k + 1}{k + 1}, \frac{2 k + 4}{k + 1}, \frac{10 k + 6}{k + 1})$
So,
$\frac{2 k + 4}{k + 1} = 8$
$\Rightarrow 2 k + 4 = 8 k + 8$
$\Rightarrow 6 k = - 4$
$k = - 2 / 3$
And,
$\frac{5 k + 1}{k + 1} = a$
$\Rightarrow \frac{5 (- 2 / 3) + 1}{- 2 / 3 + 1} = a$
$\Rightarrow \frac{- 10}{3} + 1 = \frac{a}{3}$
$\Rightarrow \frac{- 7}{3} = \frac{a}{3}$
$\Rightarrow a = - 7$
Hence, value of
$\frac{2 m}{n} - \frac{a}{3}$
$= - 2 \times \frac{2}{3} + \frac{7}{3}$
$= - \frac{4}{3} + \frac{7}{3}$
$= \frac{3}{3} = 1$ Hence, $\frac{2 m}{n} - \frac{a}{3} = 1$

AP EAMCET-04.07.2021

Three Dimensional Geometry

121379 Let the image of the point $P (1, 2, 6)$ in the plane passing through the points $A (1, 2, 0), B (1, 4, 1)$ and $C (0, 5, 1)$ be $Q (α^{2} + β^{2} + γ^{2})$ is equal to :

1 65

2 70

3 76

4 62

Explanation:

A Given,
$P (1, 2, 6) & A (1, 2, 0), B (1, 4, 1), C (0, 5, 1)$
Equation of required plane will be
$a (x - 1) + b (y - 2) + c (z - 0) = 0$
Putting $(1, 4, 1) \Rightarrow 2 b + c = 0$
putting $(0, 5, 1) \Rightarrow - a + 3 b + c = 0$
On solving (i) \& (ii) we get -
$a = b and c = - 2 b$
So,
$1 (x - 1) + 1 (y - 2) - 2 (z - 0) = 0$
$\Rightarrow x + y - 2 z - 3 = 0$
Image of $P$ on this plane is -
$\frac{x - 1}{1} = \frac{β - 2}{1} = \frac{γ - 6}{- 2} = \frac{- 2 (1 + 2 - 12 - 3)}{6}$
$\Rightarrow \frac{α - 1}{1} = \frac{β - 2}{1} = \frac{γ - 6}{- 2} = 4$
$\Rightarrow α = 5, β = 6, γ = - 2$
$So, α^{2} + β^{2} + γ^{2} = 25 + 36 + 4$
$\Rightarrow α^{2} + β^{2} + γ^{2} = 65$

JEE Main-10.04.2023

Three Dimensional Geometry

121381 The perpendicular distance from origin to the plane $x + 2 y - 2 z + 5 = 0$ equals units.

1

\frac{3}{5}

2

\frac{5}{3}

3

\frac{5}{9}

4 5

Explanation:

B
Distance from $(0, 0)$ to the plane $x + 2 y - 2 z + 5 = 0$ is
$d = | \frac{a x_{1} + b y_{1} + c z_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} |$
$= | \frac{1 (0) + 2 (0) - 2 (0) + 5}{\sqrt{1^{2} + 2^{2} + (- 2)^{2}}} |$
$\Rightarrow d = \frac{5}{3}$

APEAPCET-20.08.2021

Three Dimensional Geometry

121382 If $(3, 4, - 7)$ is the foot of the perpendicular drawn from the point $(- 2, 3, 6)$ to the plane $π$ then the sum of the intercepts made by the plane $π$ on the $x$ and $y$ -axes is

1 132

2 142

3 210

4 175

Explanation:

A Given,
$A (3, 4, - 7), B (- 2, 3, 6)$
$\Rightarrow \vec{AB} = - 5 \hat{i} - \hat{j} + 13 \hat{k}$
Equation of plane $λ$ containing point $A$ having normal
vector $\vec{AB}$ is
$λ : - 5 (x - 3) - 1 (y - 4) + 13 (z + 7) = 0$
$\Rightarrow - 5 x - y + 13 z = - 110$
Or $5 x + y - 13 z = 110$
$\frac{x}{22} + \frac{y}{110} - \frac{z}{110 / 13} = 1$
Sum of intercept on $x$ and $y$ axes is
$22 + 110 = 132$

AP EAMCET-06.07.2022

Three Dimensional Geometry

121383 If the point $(a, 8, - 2)$ divides the line segment joining the points $(1, 4, 6)$ and $(5, 2, 10)$ in the ratio $m$ : $n$ then $\frac{2 m}{n} - \frac{a}{3} =$

1 -7

2 1

3 -2

4 3

Explanation:

B Given, points
$A (a, 8, - 2), P (1, 4, 6)$ and $Q (5, 2, 10)$ where, ' $A$ ' divides $PQ$ is $k : 1$ ratio.
Then,
Appling section formula -
$A = (\frac{k \times 5 + 1 \times 1}{k + 1}, \frac{k \times 2 + 1 \times 4}{k + 1}, \frac{k \times 10 + 1 \times 6}{k + 1})$
$\Rightarrow (a, 8, - 2) = (\frac{5 k + 1}{k + 1}, \frac{2 k + 4}{k + 1}, \frac{10 k + 6}{k + 1})$
So,
$\frac{2 k + 4}{k + 1} = 8$
$\Rightarrow 2 k + 4 = 8 k + 8$
$\Rightarrow 6 k = - 4$
$k = - 2 / 3$
And,
$\frac{5 k + 1}{k + 1} = a$
$\Rightarrow \frac{5 (- 2 / 3) + 1}{- 2 / 3 + 1} = a$
$\Rightarrow \frac{- 10}{3} + 1 = \frac{a}{3}$
$\Rightarrow \frac{- 7}{3} = \frac{a}{3}$
$\Rightarrow a = - 7$
Hence, value of
$\frac{2 m}{n} - \frac{a}{3}$
$= - 2 \times \frac{2}{3} + \frac{7}{3}$
$= - \frac{4}{3} + \frac{7}{3}$
$= \frac{3}{3} = 1$ Hence, $\frac{2 m}{n} - \frac{a}{3} = 1$

AP EAMCET-04.07.2021

Three Dimensional Geometry

121379 Let the image of the point $P (1, 2, 6)$ in the plane passing through the points $A (1, 2, 0), B (1, 4, 1)$ and $C (0, 5, 1)$ be $Q (α^{2} + β^{2} + γ^{2})$ is equal to :

1 65

2 70

3 76

4 62

Explanation:

A Given,
$P (1, 2, 6) & A (1, 2, 0), B (1, 4, 1), C (0, 5, 1)$
Equation of required plane will be
$a (x - 1) + b (y - 2) + c (z - 0) = 0$
Putting $(1, 4, 1) \Rightarrow 2 b + c = 0$
putting $(0, 5, 1) \Rightarrow - a + 3 b + c = 0$
On solving (i) \& (ii) we get -
$a = b and c = - 2 b$
So,
$1 (x - 1) + 1 (y - 2) - 2 (z - 0) = 0$
$\Rightarrow x + y - 2 z - 3 = 0$
Image of $P$ on this plane is -
$\frac{x - 1}{1} = \frac{β - 2}{1} = \frac{γ - 6}{- 2} = \frac{- 2 (1 + 2 - 12 - 3)}{6}$
$\Rightarrow \frac{α - 1}{1} = \frac{β - 2}{1} = \frac{γ - 6}{- 2} = 4$
$\Rightarrow α = 5, β = 6, γ = - 2$
$So, α^{2} + β^{2} + γ^{2} = 25 + 36 + 4$
$\Rightarrow α^{2} + β^{2} + γ^{2} = 65$

JEE Main-10.04.2023

Three Dimensional Geometry

121381 The perpendicular distance from origin to the plane $x + 2 y - 2 z + 5 = 0$ equals units.

1

\frac{3}{5}

2

\frac{5}{3}

3

\frac{5}{9}

4 5

Explanation:

B
Distance from $(0, 0)$ to the plane $x + 2 y - 2 z + 5 = 0$ is
$d = | \frac{a x_{1} + b y_{1} + c z_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} |$
$= | \frac{1 (0) + 2 (0) - 2 (0) + 5}{\sqrt{1^{2} + 2^{2} + (- 2)^{2}}} |$
$\Rightarrow d = \frac{5}{3}$

APEAPCET-20.08.2021

Three Dimensional Geometry

121382 If $(3, 4, - 7)$ is the foot of the perpendicular drawn from the point $(- 2, 3, 6)$ to the plane $π$ then the sum of the intercepts made by the plane $π$ on the $x$ and $y$ -axes is

1 132

2 142

3 210

4 175

Explanation:

A Given,
$A (3, 4, - 7), B (- 2, 3, 6)$
$\Rightarrow \vec{AB} = - 5 \hat{i} - \hat{j} + 13 \hat{k}$
Equation of plane $λ$ containing point $A$ having normal
vector $\vec{AB}$ is
$λ : - 5 (x - 3) - 1 (y - 4) + 13 (z + 7) = 0$
$\Rightarrow - 5 x - y + 13 z = - 110$
Or $5 x + y - 13 z = 110$
$\frac{x}{22} + \frac{y}{110} - \frac{z}{110 / 13} = 1$
Sum of intercept on $x$ and $y$ axes is
$22 + 110 = 132$

AP EAMCET-06.07.2022

Three Dimensional Geometry

121383 If the point $(a, 8, - 2)$ divides the line segment joining the points $(1, 4, 6)$ and $(5, 2, 10)$ in the ratio $m$ : $n$ then $\frac{2 m}{n} - \frac{a}{3} =$

1 -7

2 1

3 -2

4 3

Explanation:

B Given, points
$A (a, 8, - 2), P (1, 4, 6)$ and $Q (5, 2, 10)$ where, ' $A$ ' divides $PQ$ is $k : 1$ ratio.
Then,
Appling section formula -
$A = (\frac{k \times 5 + 1 \times 1}{k + 1}, \frac{k \times 2 + 1 \times 4}{k + 1}, \frac{k \times 10 + 1 \times 6}{k + 1})$
$\Rightarrow (a, 8, - 2) = (\frac{5 k + 1}{k + 1}, \frac{2 k + 4}{k + 1}, \frac{10 k + 6}{k + 1})$
So,
$\frac{2 k + 4}{k + 1} = 8$
$\Rightarrow 2 k + 4 = 8 k + 8$
$\Rightarrow 6 k = - 4$
$k = - 2 / 3$
And,
$\frac{5 k + 1}{k + 1} = a$
$\Rightarrow \frac{5 (- 2 / 3) + 1}{- 2 / 3 + 1} = a$
$\Rightarrow \frac{- 10}{3} + 1 = \frac{a}{3}$
$\Rightarrow \frac{- 7}{3} = \frac{a}{3}$
$\Rightarrow a = - 7$
Hence, value of
$\frac{2 m}{n} - \frac{a}{3}$
$= - 2 \times \frac{2}{3} + \frac{7}{3}$
$= - \frac{4}{3} + \frac{7}{3}$
$= \frac{3}{3} = 1$ Hence, $\frac{2 m}{n} - \frac{a}{3} = 1$

AP EAMCET-04.07.2021

Three Dimensional Geometry

121379 Let the image of the point $P (1, 2, 6)$ in the plane passing through the points $A (1, 2, 0), B (1, 4, 1)$ and $C (0, 5, 1)$ be $Q (α^{2} + β^{2} + γ^{2})$ is equal to :

1 65

2 70

3 76

4 62

Explanation:

A Given,
$P (1, 2, 6) & A (1, 2, 0), B (1, 4, 1), C (0, 5, 1)$
Equation of required plane will be
$a (x - 1) + b (y - 2) + c (z - 0) = 0$
Putting $(1, 4, 1) \Rightarrow 2 b + c = 0$
putting $(0, 5, 1) \Rightarrow - a + 3 b + c = 0$
On solving (i) \& (ii) we get -
$a = b and c = - 2 b$
So,
$1 (x - 1) + 1 (y - 2) - 2 (z - 0) = 0$
$\Rightarrow x + y - 2 z - 3 = 0$
Image of $P$ on this plane is -
$\frac{x - 1}{1} = \frac{β - 2}{1} = \frac{γ - 6}{- 2} = \frac{- 2 (1 + 2 - 12 - 3)}{6}$
$\Rightarrow \frac{α - 1}{1} = \frac{β - 2}{1} = \frac{γ - 6}{- 2} = 4$
$\Rightarrow α = 5, β = 6, γ = - 2$
$So, α^{2} + β^{2} + γ^{2} = 25 + 36 + 4$
$\Rightarrow α^{2} + β^{2} + γ^{2} = 65$

JEE Main-10.04.2023

Three Dimensional Geometry

121381 The perpendicular distance from origin to the plane $x + 2 y - 2 z + 5 = 0$ equals units.

1

\frac{3}{5}

2

\frac{5}{3}

3

\frac{5}{9}

4 5

Explanation:

B
Distance from $(0, 0)$ to the plane $x + 2 y - 2 z + 5 = 0$ is
$d = | \frac{a x_{1} + b y_{1} + c z_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} |$
$= | \frac{1 (0) + 2 (0) - 2 (0) + 5}{\sqrt{1^{2} + 2^{2} + (- 2)^{2}}} |$
$\Rightarrow d = \frac{5}{3}$

APEAPCET-20.08.2021

Three Dimensional Geometry

121382 If $(3, 4, - 7)$ is the foot of the perpendicular drawn from the point $(- 2, 3, 6)$ to the plane $π$ then the sum of the intercepts made by the plane $π$ on the $x$ and $y$ -axes is

1 132

2 142

3 210

4 175

Explanation:

A Given,
$A (3, 4, - 7), B (- 2, 3, 6)$
$\Rightarrow \vec{AB} = - 5 \hat{i} - \hat{j} + 13 \hat{k}$
Equation of plane $λ$ containing point $A$ having normal
vector $\vec{AB}$ is
$λ : - 5 (x - 3) - 1 (y - 4) + 13 (z + 7) = 0$
$\Rightarrow - 5 x - y + 13 z = - 110$
Or $5 x + y - 13 z = 110$
$\frac{x}{22} + \frac{y}{110} - \frac{z}{110 / 13} = 1$
Sum of intercept on $x$ and $y$ axes is
$22 + 110 = 132$

AP EAMCET-06.07.2022

Three Dimensional Geometry

121383 If the point $(a, 8, - 2)$ divides the line segment joining the points $(1, 4, 6)$ and $(5, 2, 10)$ in the ratio $m$ : $n$ then $\frac{2 m}{n} - \frac{a}{3} =$

1 -7

2 1

3 -2

4 3

Explanation:

B Given, points
$A (a, 8, - 2), P (1, 4, 6)$ and $Q (5, 2, 10)$ where, ' $A$ ' divides $PQ$ is $k : 1$ ratio.
Then,
Appling section formula -
$A = (\frac{k \times 5 + 1 \times 1}{k + 1}, \frac{k \times 2 + 1 \times 4}{k + 1}, \frac{k \times 10 + 1 \times 6}{k + 1})$
$\Rightarrow (a, 8, - 2) = (\frac{5 k + 1}{k + 1}, \frac{2 k + 4}{k + 1}, \frac{10 k + 6}{k + 1})$
So,
$\frac{2 k + 4}{k + 1} = 8$
$\Rightarrow 2 k + 4 = 8 k + 8$
$\Rightarrow 6 k = - 4$
$k = - 2 / 3$
And,
$\frac{5 k + 1}{k + 1} = a$
$\Rightarrow \frac{5 (- 2 / 3) + 1}{- 2 / 3 + 1} = a$
$\Rightarrow \frac{- 10}{3} + 1 = \frac{a}{3}$
$\Rightarrow \frac{- 7}{3} = \frac{a}{3}$
$\Rightarrow a = - 7$
Hence, value of
$\frac{2 m}{n} - \frac{a}{3}$
$= - 2 \times \frac{2}{3} + \frac{7}{3}$
$= - \frac{4}{3} + \frac{7}{3}$
$= \frac{3}{3} = 1$ Hence, $\frac{2 m}{n} - \frac{a}{3} = 1$

AP EAMCET-04.07.2021

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