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Three Dimensional Geometry

121396 If a plane $π$ passes through the point $(- 1, 6, 2)$ is perpendicular to the planes $x + 2 y + 2 z - 5 =$ 0 and $3 x + 3 y + 2 z - 8 = 0$ , then, the perpendicular distance from the point $(1, - 1, 1)$ to the plane $π$ is

1

\frac{20}{\sqrt{29}}

2

\frac{21}{\sqrt{29}}

3

\frac{27}{\sqrt{29}}

4

\sqrt{29}

Explanation:

D Equation of plane passing through the point
$(- 1, 6, 2)$ and perpendicular to planes
$x + 2 y + 2 z - 5 = 0$ and $3 x + 3 y + 2 z - 8 = 0$ is
$| \begin{array}{ccc} x + 1 & y - 6 & z - 2 \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{array} | = 0$
$\Rightarrow (x + 1) (4 - 6) - (y - 6) (2 - 6) + (z - 2) (3 - 6) = 0$
$\Rightarrow - 2 x - 2 + 4 y - 24 - 3 z + 6 = 0$
$\Rightarrow 2 x - 4 y + 3 z + 20 = 0$
$\Rightarrow$ Distance from $(1, - 1, 1)$ to plane is
$d = | \frac{2 + 4 + 3 + 20}{\sqrt{2^{2} + 4^{2} + 3^{2}}} | = \frac{29}{\sqrt{29}} = \sqrt{29}$

TS EAMCET-14.09.2020

Three Dimensional Geometry

121355 The length of the perpendicular to the plane $\vec{r} (\hat{i} - 2 \hat{j} + 3 \hat{k}) = 14$ from the origin is

1 7 units

2

\sqrt{14}

units

3 14 units

4

\sqrt{7}

units

Explanation:

B
Length of perpendicular from the point $A (\vec{a})$ to the Plane $\vec{r} . \vec{n} = p$ is $\frac{| \vec{a} \cdot \vec{n} - p |}{| \vec{n} |}$
Here, $\vec{a} . \vec{n} = 0$ and $| \vec{n} | = \sqrt{1 + 4 + 9} = \sqrt{14}$
Hence, required distance is $\frac{| 0 - 14 |}{\sqrt{14}} = \sqrt{14}$

MHT CET-2020

Three Dimensional Geometry

121357 The unit vector perpendicular to the plane $4 x - 3 y + 12 z = 15$ is

1

\frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}

2

\frac{- 4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}

3

\frac{4 \hat{i} + 3 \hat{j} + 12 \hat{k}}{13}

4

\frac{- 4 \hat{i} + 3 \hat{j} + 12 \hat{k}}{13}

Explanation:

A Given,
The unit vector perpendicular to plane $4 x - 3 y + 12 z = 15$ is
$\frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{\sqrt{16 + 9 + 144}} = \frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}$

MHT CET-2020

Three Dimensional Geometry

121361 Distance of the point $(2, 3, 4)$ from the plane $3 x - 6 y + 2 z + 11 = 0$ is

1 0

2 1

3 2

4 3

Explanation:

B The distance of a point $P (x_{1}, y_{1}, z_{1})$ from a plane $ax + by + cz + d = 0$ is
$d = | \frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} |$
$∴$ Distance of the point $(2, 3, 4)$ from the plane $3 x - 6 y + 2 z + 11 = 0$ is
$\Rightarrow d = | \frac{6 - 18 + 8 + 11}{\sqrt{9 + 36 + 4}} | = | \frac{7}{\sqrt{49}} | = | \frac{7}{7} |$

CG PET- 2015

Three Dimensional Geometry

121396 If a plane $π$ passes through the point $(- 1, 6, 2)$ is perpendicular to the planes $x + 2 y + 2 z - 5 =$ 0 and $3 x + 3 y + 2 z - 8 = 0$ , then, the perpendicular distance from the point $(1, - 1, 1)$ to the plane $π$ is

1

\frac{20}{\sqrt{29}}

2

\frac{21}{\sqrt{29}}

3

\frac{27}{\sqrt{29}}

4

\sqrt{29}

Explanation:

D Equation of plane passing through the point
$(- 1, 6, 2)$ and perpendicular to planes
$x + 2 y + 2 z - 5 = 0$ and $3 x + 3 y + 2 z - 8 = 0$ is
$| \begin{array}{ccc} x + 1 & y - 6 & z - 2 \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{array} | = 0$
$\Rightarrow (x + 1) (4 - 6) - (y - 6) (2 - 6) + (z - 2) (3 - 6) = 0$
$\Rightarrow - 2 x - 2 + 4 y - 24 - 3 z + 6 = 0$
$\Rightarrow 2 x - 4 y + 3 z + 20 = 0$
$\Rightarrow$ Distance from $(1, - 1, 1)$ to plane is
$d = | \frac{2 + 4 + 3 + 20}{\sqrt{2^{2} + 4^{2} + 3^{2}}} | = \frac{29}{\sqrt{29}} = \sqrt{29}$

TS EAMCET-14.09.2020

Three Dimensional Geometry

121355 The length of the perpendicular to the plane $\vec{r} (\hat{i} - 2 \hat{j} + 3 \hat{k}) = 14$ from the origin is

1 7 units

2

\sqrt{14}

units

3 14 units

4

\sqrt{7}

units

Explanation:

B
Length of perpendicular from the point $A (\vec{a})$ to the Plane $\vec{r} . \vec{n} = p$ is $\frac{| \vec{a} \cdot \vec{n} - p |}{| \vec{n} |}$
Here, $\vec{a} . \vec{n} = 0$ and $| \vec{n} | = \sqrt{1 + 4 + 9} = \sqrt{14}$
Hence, required distance is $\frac{| 0 - 14 |}{\sqrt{14}} = \sqrt{14}$

MHT CET-2020

Three Dimensional Geometry

121357 The unit vector perpendicular to the plane $4 x - 3 y + 12 z = 15$ is

1

\frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}

2

\frac{- 4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}

3

\frac{4 \hat{i} + 3 \hat{j} + 12 \hat{k}}{13}

4

\frac{- 4 \hat{i} + 3 \hat{j} + 12 \hat{k}}{13}

Explanation:

A Given,
The unit vector perpendicular to plane $4 x - 3 y + 12 z = 15$ is
$\frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{\sqrt{16 + 9 + 144}} = \frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}$

MHT CET-2020

Three Dimensional Geometry

121361 Distance of the point $(2, 3, 4)$ from the plane $3 x - 6 y + 2 z + 11 = 0$ is

1 0

2 1

3 2

4 3

Explanation:

B The distance of a point $P (x_{1}, y_{1}, z_{1})$ from a plane $ax + by + cz + d = 0$ is
$d = | \frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} |$
$∴$ Distance of the point $(2, 3, 4)$ from the plane $3 x - 6 y + 2 z + 11 = 0$ is
$\Rightarrow d = | \frac{6 - 18 + 8 + 11}{\sqrt{9 + 36 + 4}} | = | \frac{7}{\sqrt{49}} | = | \frac{7}{7} |$

CG PET- 2015

Three Dimensional Geometry

121396 If a plane $π$ passes through the point $(- 1, 6, 2)$ is perpendicular to the planes $x + 2 y + 2 z - 5 =$ 0 and $3 x + 3 y + 2 z - 8 = 0$ , then, the perpendicular distance from the point $(1, - 1, 1)$ to the plane $π$ is

1

\frac{20}{\sqrt{29}}

2

\frac{21}{\sqrt{29}}

3

\frac{27}{\sqrt{29}}

4

\sqrt{29}

Explanation:

D Equation of plane passing through the point
$(- 1, 6, 2)$ and perpendicular to planes
$x + 2 y + 2 z - 5 = 0$ and $3 x + 3 y + 2 z - 8 = 0$ is
$| \begin{array}{ccc} x + 1 & y - 6 & z - 2 \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{array} | = 0$
$\Rightarrow (x + 1) (4 - 6) - (y - 6) (2 - 6) + (z - 2) (3 - 6) = 0$
$\Rightarrow - 2 x - 2 + 4 y - 24 - 3 z + 6 = 0$
$\Rightarrow 2 x - 4 y + 3 z + 20 = 0$
$\Rightarrow$ Distance from $(1, - 1, 1)$ to plane is
$d = | \frac{2 + 4 + 3 + 20}{\sqrt{2^{2} + 4^{2} + 3^{2}}} | = \frac{29}{\sqrt{29}} = \sqrt{29}$

TS EAMCET-14.09.2020

Three Dimensional Geometry

121355 The length of the perpendicular to the plane $\vec{r} (\hat{i} - 2 \hat{j} + 3 \hat{k}) = 14$ from the origin is

1 7 units

2

\sqrt{14}

units

3 14 units

4

\sqrt{7}

units

Explanation:

B
Length of perpendicular from the point $A (\vec{a})$ to the Plane $\vec{r} . \vec{n} = p$ is $\frac{| \vec{a} \cdot \vec{n} - p |}{| \vec{n} |}$
Here, $\vec{a} . \vec{n} = 0$ and $| \vec{n} | = \sqrt{1 + 4 + 9} = \sqrt{14}$
Hence, required distance is $\frac{| 0 - 14 |}{\sqrt{14}} = \sqrt{14}$

MHT CET-2020

Three Dimensional Geometry

121357 The unit vector perpendicular to the plane $4 x - 3 y + 12 z = 15$ is

1

\frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}

2

\frac{- 4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}

3

\frac{4 \hat{i} + 3 \hat{j} + 12 \hat{k}}{13}

4

\frac{- 4 \hat{i} + 3 \hat{j} + 12 \hat{k}}{13}

Explanation:

A Given,
The unit vector perpendicular to plane $4 x - 3 y + 12 z = 15$ is
$\frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{\sqrt{16 + 9 + 144}} = \frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}$

MHT CET-2020

Three Dimensional Geometry

121361 Distance of the point $(2, 3, 4)$ from the plane $3 x - 6 y + 2 z + 11 = 0$ is

1 0

2 1

3 2

4 3

Explanation:

B The distance of a point $P (x_{1}, y_{1}, z_{1})$ from a plane $ax + by + cz + d = 0$ is
$d = | \frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} |$
$∴$ Distance of the point $(2, 3, 4)$ from the plane $3 x - 6 y + 2 z + 11 = 0$ is
$\Rightarrow d = | \frac{6 - 18 + 8 + 11}{\sqrt{9 + 36 + 4}} | = | \frac{7}{\sqrt{49}} | = | \frac{7}{7} |$

CG PET- 2015

Three Dimensional Geometry

121396 If a plane $π$ passes through the point $(- 1, 6, 2)$ is perpendicular to the planes $x + 2 y + 2 z - 5 =$ 0 and $3 x + 3 y + 2 z - 8 = 0$ , then, the perpendicular distance from the point $(1, - 1, 1)$ to the plane $π$ is

1

\frac{20}{\sqrt{29}}

2

\frac{21}{\sqrt{29}}

3

\frac{27}{\sqrt{29}}

4

\sqrt{29}

Explanation:

D Equation of plane passing through the point
$(- 1, 6, 2)$ and perpendicular to planes
$x + 2 y + 2 z - 5 = 0$ and $3 x + 3 y + 2 z - 8 = 0$ is
$| \begin{array}{ccc} x + 1 & y - 6 & z - 2 \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{array} | = 0$
$\Rightarrow (x + 1) (4 - 6) - (y - 6) (2 - 6) + (z - 2) (3 - 6) = 0$
$\Rightarrow - 2 x - 2 + 4 y - 24 - 3 z + 6 = 0$
$\Rightarrow 2 x - 4 y + 3 z + 20 = 0$
$\Rightarrow$ Distance from $(1, - 1, 1)$ to plane is
$d = | \frac{2 + 4 + 3 + 20}{\sqrt{2^{2} + 4^{2} + 3^{2}}} | = \frac{29}{\sqrt{29}} = \sqrt{29}$

TS EAMCET-14.09.2020

Three Dimensional Geometry

121355 The length of the perpendicular to the plane $\vec{r} (\hat{i} - 2 \hat{j} + 3 \hat{k}) = 14$ from the origin is

1 7 units

2

\sqrt{14}

units

3 14 units

4

\sqrt{7}

units

Explanation:

B
Length of perpendicular from the point $A (\vec{a})$ to the Plane $\vec{r} . \vec{n} = p$ is $\frac{| \vec{a} \cdot \vec{n} - p |}{| \vec{n} |}$
Here, $\vec{a} . \vec{n} = 0$ and $| \vec{n} | = \sqrt{1 + 4 + 9} = \sqrt{14}$
Hence, required distance is $\frac{| 0 - 14 |}{\sqrt{14}} = \sqrt{14}$

MHT CET-2020

Three Dimensional Geometry

121357 The unit vector perpendicular to the plane $4 x - 3 y + 12 z = 15$ is

1

\frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}

2

\frac{- 4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}

3

\frac{4 \hat{i} + 3 \hat{j} + 12 \hat{k}}{13}

4

\frac{- 4 \hat{i} + 3 \hat{j} + 12 \hat{k}}{13}

Explanation:

A Given,
The unit vector perpendicular to plane $4 x - 3 y + 12 z = 15$ is
$\frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{\sqrt{16 + 9 + 144}} = \frac{4 \hat{i} - 3 \hat{j} + 12 \hat{k}}{13}$

MHT CET-2020

Three Dimensional Geometry

121361 Distance of the point $(2, 3, 4)$ from the plane $3 x - 6 y + 2 z + 11 = 0$ is

1 0

2 1

3 2

4 3

Explanation:

B The distance of a point $P (x_{1}, y_{1}, z_{1})$ from a plane $ax + by + cz + d = 0$ is
$d = | \frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}} |$
$∴$ Distance of the point $(2, 3, 4)$ from the plane $3 x - 6 y + 2 z + 11 = 0$ is
$\Rightarrow d = | \frac{6 - 18 + 8 + 11}{\sqrt{9 + 36 + 4}} | = | \frac{7}{\sqrt{49}} | = | \frac{7}{7} |$

CG PET- 2015

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