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Three Dimensional Geometry

121431 A variable plane passes through a fixed point $(α, β, γ)$ and meets the coordinate axes in $A, B$ and $C$ . Let $P_{1}, P_{2}$ and $P_{3}$ be the planes passing through $A, B, C$ and parallel to the coordinate planes YZ, ZX, XY respectively. Then, the locus of the point of intersection of the planes $P_{1}, P_{2}$ and $P_{3}$ is

1

α x + β y + γ z = 1

2

\frac{α}{x} + \frac{β}{y} + \frac{γ}{z} = 1

3

α x^{2} + β y^{2} + γ z^{2} = 1

4

α β x + β γ y + α γ z = 1

Explanation:

B Given,
Point $(α, β, γ)$
Let the equation of required plane is -
$\frac{x}{x_{0}} + \frac{y}{y_{0}} + \frac{z}{z_{0}} = 1$
Since, this plane passes through given point $(α, β, γ)$
Hence,
$\frac{α}{x_{0}} + \frac{β}{y_{0}} + \frac{γ}{z_{0}} = 1$
$∴$ Locus of $(x_{0}, y_{0}, z_{0})$ is $\frac{α}{x} + \frac{β}{y} + \frac{γ}{z} = 1$

AP EAMCET-22.04.2018

Three Dimensional Geometry

121433 The perimeter of the triangle with vertices at $(1, 0, 0), (0, 1, 0)$ and $(0, 0, 1)$ is

1 3

2 2

3

2 \sqrt{2}

4

3 \sqrt{2}

Explanation:

D Given,
$A (1, 0, 0), B (0, 1, 0)$ and $c (0, 0, 1)$
$\Rightarrow \vec{AB} = - \hat{i} + \hat{j}$
\& $| \vec{AB} | = \sqrt{1 + 1} = \sqrt{2}$
$\vec{BC} = - \hat{j} + \hat{k}$
\& $| \vec{BC} | = \sqrt{1 + 1} = \sqrt{2}$
$\vec{CA} = \hat{i} + \hat{k}$
$| \vec{CA} | = \sqrt{1 + 1} = \sqrt{2}$
Hence, perimeter of $△ ABC = AB + BC + CA$
$= 3 \sqrt{2}$

AP EAMCET-2009

Three Dimensional Geometry

121434 A variable plane passes through a fixed point $(1, 2, 3)$ Then, the foot of the perpendicular from the origin to the plane lies on

1 a circle

2 a sphere

3 an ellipse

4 a parabola

Explanation:

B Given, Point $(1, 2, 3)$
Equation of required plane
$a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0$
$\Rightarrow a (1 - x_{0}) + b (2 - y_{0}) + c (3 - z_{0}) = 0$
Or
$a (1 - a) + b (2 - b) + c (3 - c) = 0$
$\Rightarrow a^{2} + b^{2} + c^{2} - a - 2 b - 2 c = 0$
or
$x^{2} + y^{2} + z^{2} - x - 2 y - 2 z = 0$
Here, we can see that this is the equation of a sphere.

AP EAMCET-2013

Three Dimensional Geometry

121435 If the plane $3 x - 2 y - z - 18 = 0$ meets the coordinate axes in $A, B, C$ then the centroid of $△ ABC$ is

1

(2, 3, - 6)

2

(2, - 3, 6)

3

(- 2, - 3, 6)

4

(2, - 3, - 6)

Explanation:

D Given,
Plane
P : $3 x - 2 y - z - 18 = 0$
Or
$\frac{x}{6} - \frac{y}{9} - \frac{z}{18} = 1$
Now $A = (6, 0, 0), B (0, - 9, 0)$ and $C = (0, 0, - 18)$
Hence, centroid of $△ ABC$
$= (\frac{6 + 0 + 0}{3}, \frac{0 - 9 + 0}{3}, \frac{0 + 0 - 18}{3})$
$\Rightarrow Centroid of △ ABC = [\frac{6}{3}, \frac{- 9}{3}, \frac{- 18}{3}]$
$= (2, - 3, - 6)$

AP EAMCET-2004

Three Dimensional Geometry

121431 A variable plane passes through a fixed point $(α, β, γ)$ and meets the coordinate axes in $A, B$ and $C$ . Let $P_{1}, P_{2}$ and $P_{3}$ be the planes passing through $A, B, C$ and parallel to the coordinate planes YZ, ZX, XY respectively. Then, the locus of the point of intersection of the planes $P_{1}, P_{2}$ and $P_{3}$ is

1

α x + β y + γ z = 1

2

\frac{α}{x} + \frac{β}{y} + \frac{γ}{z} = 1

3

α x^{2} + β y^{2} + γ z^{2} = 1

4

α β x + β γ y + α γ z = 1

Explanation:

B Given,
Point $(α, β, γ)$
Let the equation of required plane is -
$\frac{x}{x_{0}} + \frac{y}{y_{0}} + \frac{z}{z_{0}} = 1$
Since, this plane passes through given point $(α, β, γ)$
Hence,
$\frac{α}{x_{0}} + \frac{β}{y_{0}} + \frac{γ}{z_{0}} = 1$
$∴$ Locus of $(x_{0}, y_{0}, z_{0})$ is $\frac{α}{x} + \frac{β}{y} + \frac{γ}{z} = 1$

AP EAMCET-22.04.2018

Three Dimensional Geometry

121433 The perimeter of the triangle with vertices at $(1, 0, 0), (0, 1, 0)$ and $(0, 0, 1)$ is

1 3

2 2

3

2 \sqrt{2}

4

3 \sqrt{2}

Explanation:

D Given,
$A (1, 0, 0), B (0, 1, 0)$ and $c (0, 0, 1)$
$\Rightarrow \vec{AB} = - \hat{i} + \hat{j}$
\& $| \vec{AB} | = \sqrt{1 + 1} = \sqrt{2}$
$\vec{BC} = - \hat{j} + \hat{k}$
\& $| \vec{BC} | = \sqrt{1 + 1} = \sqrt{2}$
$\vec{CA} = \hat{i} + \hat{k}$
$| \vec{CA} | = \sqrt{1 + 1} = \sqrt{2}$
Hence, perimeter of $△ ABC = AB + BC + CA$
$= 3 \sqrt{2}$

AP EAMCET-2009

Three Dimensional Geometry

121434 A variable plane passes through a fixed point $(1, 2, 3)$ Then, the foot of the perpendicular from the origin to the plane lies on

1 a circle

2 a sphere

3 an ellipse

4 a parabola

Explanation:

B Given, Point $(1, 2, 3)$
Equation of required plane
$a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0$
$\Rightarrow a (1 - x_{0}) + b (2 - y_{0}) + c (3 - z_{0}) = 0$
Or
$a (1 - a) + b (2 - b) + c (3 - c) = 0$
$\Rightarrow a^{2} + b^{2} + c^{2} - a - 2 b - 2 c = 0$
or
$x^{2} + y^{2} + z^{2} - x - 2 y - 2 z = 0$
Here, we can see that this is the equation of a sphere.

AP EAMCET-2013

Three Dimensional Geometry

121435 If the plane $3 x - 2 y - z - 18 = 0$ meets the coordinate axes in $A, B, C$ then the centroid of $△ ABC$ is

1

(2, 3, - 6)

2

(2, - 3, 6)

3

(- 2, - 3, 6)

4

(2, - 3, - 6)

Explanation:

D Given,
Plane
P : $3 x - 2 y - z - 18 = 0$
Or
$\frac{x}{6} - \frac{y}{9} - \frac{z}{18} = 1$
Now $A = (6, 0, 0), B (0, - 9, 0)$ and $C = (0, 0, - 18)$
Hence, centroid of $△ ABC$
$= (\frac{6 + 0 + 0}{3}, \frac{0 - 9 + 0}{3}, \frac{0 + 0 - 18}{3})$
$\Rightarrow Centroid of △ ABC = [\frac{6}{3}, \frac{- 9}{3}, \frac{- 18}{3}]$
$= (2, - 3, - 6)$

AP EAMCET-2004

Three Dimensional Geometry

121431 A variable plane passes through a fixed point $(α, β, γ)$ and meets the coordinate axes in $A, B$ and $C$ . Let $P_{1}, P_{2}$ and $P_{3}$ be the planes passing through $A, B, C$ and parallel to the coordinate planes YZ, ZX, XY respectively. Then, the locus of the point of intersection of the planes $P_{1}, P_{2}$ and $P_{3}$ is

1

α x + β y + γ z = 1

2

\frac{α}{x} + \frac{β}{y} + \frac{γ}{z} = 1

3

α x^{2} + β y^{2} + γ z^{2} = 1

4

α β x + β γ y + α γ z = 1

Explanation:

B Given,
Point $(α, β, γ)$
Let the equation of required plane is -
$\frac{x}{x_{0}} + \frac{y}{y_{0}} + \frac{z}{z_{0}} = 1$
Since, this plane passes through given point $(α, β, γ)$
Hence,
$\frac{α}{x_{0}} + \frac{β}{y_{0}} + \frac{γ}{z_{0}} = 1$
$∴$ Locus of $(x_{0}, y_{0}, z_{0})$ is $\frac{α}{x} + \frac{β}{y} + \frac{γ}{z} = 1$

AP EAMCET-22.04.2018

Three Dimensional Geometry

121433 The perimeter of the triangle with vertices at $(1, 0, 0), (0, 1, 0)$ and $(0, 0, 1)$ is

1 3

2 2

3

2 \sqrt{2}

4

3 \sqrt{2}

Explanation:

D Given,
$A (1, 0, 0), B (0, 1, 0)$ and $c (0, 0, 1)$
$\Rightarrow \vec{AB} = - \hat{i} + \hat{j}$
\& $| \vec{AB} | = \sqrt{1 + 1} = \sqrt{2}$
$\vec{BC} = - \hat{j} + \hat{k}$
\& $| \vec{BC} | = \sqrt{1 + 1} = \sqrt{2}$
$\vec{CA} = \hat{i} + \hat{k}$
$| \vec{CA} | = \sqrt{1 + 1} = \sqrt{2}$
Hence, perimeter of $△ ABC = AB + BC + CA$
$= 3 \sqrt{2}$

AP EAMCET-2009

Three Dimensional Geometry

121434 A variable plane passes through a fixed point $(1, 2, 3)$ Then, the foot of the perpendicular from the origin to the plane lies on

1 a circle

2 a sphere

3 an ellipse

4 a parabola

Explanation:

B Given, Point $(1, 2, 3)$
Equation of required plane
$a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0$
$\Rightarrow a (1 - x_{0}) + b (2 - y_{0}) + c (3 - z_{0}) = 0$
Or
$a (1 - a) + b (2 - b) + c (3 - c) = 0$
$\Rightarrow a^{2} + b^{2} + c^{2} - a - 2 b - 2 c = 0$
or
$x^{2} + y^{2} + z^{2} - x - 2 y - 2 z = 0$
Here, we can see that this is the equation of a sphere.

AP EAMCET-2013

Three Dimensional Geometry

121435 If the plane $3 x - 2 y - z - 18 = 0$ meets the coordinate axes in $A, B, C$ then the centroid of $△ ABC$ is

1

(2, 3, - 6)

2

(2, - 3, 6)

3

(- 2, - 3, 6)

4

(2, - 3, - 6)

Explanation:

D Given,
Plane
P : $3 x - 2 y - z - 18 = 0$
Or
$\frac{x}{6} - \frac{y}{9} - \frac{z}{18} = 1$
Now $A = (6, 0, 0), B (0, - 9, 0)$ and $C = (0, 0, - 18)$
Hence, centroid of $△ ABC$
$= (\frac{6 + 0 + 0}{3}, \frac{0 - 9 + 0}{3}, \frac{0 + 0 - 18}{3})$
$\Rightarrow Centroid of △ ABC = [\frac{6}{3}, \frac{- 9}{3}, \frac{- 18}{3}]$
$= (2, - 3, - 6)$

AP EAMCET-2004

Three Dimensional Geometry

121431 A variable plane passes through a fixed point $(α, β, γ)$ and meets the coordinate axes in $A, B$ and $C$ . Let $P_{1}, P_{2}$ and $P_{3}$ be the planes passing through $A, B, C$ and parallel to the coordinate planes YZ, ZX, XY respectively. Then, the locus of the point of intersection of the planes $P_{1}, P_{2}$ and $P_{3}$ is

1

α x + β y + γ z = 1

2

\frac{α}{x} + \frac{β}{y} + \frac{γ}{z} = 1

3

α x^{2} + β y^{2} + γ z^{2} = 1

4

α β x + β γ y + α γ z = 1

Explanation:

B Given,
Point $(α, β, γ)$
Let the equation of required plane is -
$\frac{x}{x_{0}} + \frac{y}{y_{0}} + \frac{z}{z_{0}} = 1$
Since, this plane passes through given point $(α, β, γ)$
Hence,
$\frac{α}{x_{0}} + \frac{β}{y_{0}} + \frac{γ}{z_{0}} = 1$
$∴$ Locus of $(x_{0}, y_{0}, z_{0})$ is $\frac{α}{x} + \frac{β}{y} + \frac{γ}{z} = 1$

AP EAMCET-22.04.2018

Three Dimensional Geometry

121433 The perimeter of the triangle with vertices at $(1, 0, 0), (0, 1, 0)$ and $(0, 0, 1)$ is

1 3

2 2

3

2 \sqrt{2}

4

3 \sqrt{2}

Explanation:

D Given,
$A (1, 0, 0), B (0, 1, 0)$ and $c (0, 0, 1)$
$\Rightarrow \vec{AB} = - \hat{i} + \hat{j}$
\& $| \vec{AB} | = \sqrt{1 + 1} = \sqrt{2}$
$\vec{BC} = - \hat{j} + \hat{k}$
\& $| \vec{BC} | = \sqrt{1 + 1} = \sqrt{2}$
$\vec{CA} = \hat{i} + \hat{k}$
$| \vec{CA} | = \sqrt{1 + 1} = \sqrt{2}$
Hence, perimeter of $△ ABC = AB + BC + CA$
$= 3 \sqrt{2}$

AP EAMCET-2009

Three Dimensional Geometry

121434 A variable plane passes through a fixed point $(1, 2, 3)$ Then, the foot of the perpendicular from the origin to the plane lies on

1 a circle

2 a sphere

3 an ellipse

4 a parabola

Explanation:

B Given, Point $(1, 2, 3)$
Equation of required plane
$a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0$
$\Rightarrow a (1 - x_{0}) + b (2 - y_{0}) + c (3 - z_{0}) = 0$
Or
$a (1 - a) + b (2 - b) + c (3 - c) = 0$
$\Rightarrow a^{2} + b^{2} + c^{2} - a - 2 b - 2 c = 0$
or
$x^{2} + y^{2} + z^{2} - x - 2 y - 2 z = 0$
Here, we can see that this is the equation of a sphere.

AP EAMCET-2013

Three Dimensional Geometry

121435 If the plane $3 x - 2 y - z - 18 = 0$ meets the coordinate axes in $A, B, C$ then the centroid of $△ ABC$ is

1

(2, 3, - 6)

2

(2, - 3, 6)

3

(- 2, - 3, 6)

4

(2, - 3, - 6)

Explanation:

D Given,
Plane
P : $3 x - 2 y - z - 18 = 0$
Or
$\frac{x}{6} - \frac{y}{9} - \frac{z}{18} = 1$
Now $A = (6, 0, 0), B (0, - 9, 0)$ and $C = (0, 0, - 18)$
Hence, centroid of $△ ABC$
$= (\frac{6 + 0 + 0}{3}, \frac{0 - 9 + 0}{3}, \frac{0 + 0 - 18}{3})$
$\Rightarrow Centroid of △ ABC = [\frac{6}{3}, \frac{- 9}{3}, \frac{- 18}{3}]$
$= (2, - 3, - 6)$

AP EAMCET-2004