QBManager

Three Dimensional Geometry

121444 Distance between two parallel planes $2 x + y +$ $2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is

1

\frac{5}{2}

2

\frac{7}{2}

3

\frac{9}{2}

4

\frac{3}{2}

Explanation:

B The planes are
$4 x + 2 y + 4 z = 16, 4 x + 2 y + 4 z = - 5$
$\Rightarrow Distance between planes$
$= \frac{16 - (- 5)}{\sqrt{4^{2} + 2^{2} + 4^{2}}} = \frac{21}{6} = \frac{7}{2}$

COMEDK-2013

Three Dimensional Geometry

121410 From a point $P (a, b, c)$ perpendicular $P A, P B$ and drawn to $y z$ and $zx$ planes. Find the equation of the plane $O A B$ , where $O$ is the origin.

1

b c x + c a y + a b z = 0

2

b c x + c a y - a b z = 0

3

b c x - c a y + a b z = 0

4

- bcx + cay + abz = 0

Explanation:

B P(a, b, c) and PA and PB are perpendicular to $YZ$ and $ZX$ planes. Hence, co-ordinate of $A$ and $B$ are $(0, b, c)$ and $(a, 0, c)$ respectively.
Equation of plane passing through $(0, 0, 0)$ ,
$(0, b, c)$ and $(a, 0, c)$ is $| \begin{array}{lll} x & y & z \\ 0 & b & c \\ a & 0 & c \end{array} | = 0$
$\Rightarrow x (b c - 0) - y (0 - a c) + z (0 - a b) = 0$
$\Rightarrow b c x + a c y - a b z = 0$

WB JEE-2021

Three Dimensional Geometry

121404 Equation of line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the plane $x - 2 y - z + 5 = 0$ and $x + y + 3 z = 6$ is

1

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

2

\frac{x - 2}{4} = \frac{y - 3}{3} = \frac{z - 1}{2}

3

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

4

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

Explanation:

A Given
$Point (2, 3, 1)$
$P_{1} = x - 2 y - z + 5 = 0, {\vec{n}}_{1} = \hat{i} - 2 \hat{j} - \hat{k}$
$P_{2} = x + y + 3 z - 6 = 0, {\vec{n}}_{2} = \hat{i} + \hat{j} + 3 \hat{k}$
Intersection of the plane-
$Line, L = {\vec{n}}_{1} \times {\vec{n}}_{2}$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & - 1 \\ 1 & 1 & 3 \end{array} |$
$= \hat{i} (- 6 + 1) - \hat{j} (3 + 1) + \hat{k} (1 + 2) = - 5 \hat{i} - 4 \hat{j} + 3 \hat{k}$
Direction ratios of line $(- 5, - 4, 3)$ i.e. $(a, b, c)$
$∴$ Points $(2, 3, 1)$ i.e. $(x_{1}, y_{1}, z_{1})$
Equation of line, $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$
$= \frac{x - 2}{- 5} = \frac{y - 3}{- 4} = \frac{z - 1}{3}$

Karnataka CET-2015

Three Dimensional Geometry

121419 The foot of the perpendicular from the point (7, $14, 5)$ to the plane $2 x + 4 y - z = 2$ are

1

(1, 2, 8)

2

(3, 2, 8)

3

(5, 10, 6)

4

(9, 18, 4)

Explanation:

A We know that the length of the perpendicular from the point $(x_{1}, y_{1}, z_{1})$ to the plane $ax + by + cz + d = 0$ is $\frac{| {ax}_{1} + {by}_{1} + {cz}_{1} + d |}{\sqrt{a^{2} + b^{2} + c^{2}}}$ and the co-ordinate $(α, β, γ)$ or the foot of the $⊥$ are given by-
$\frac{α - x_{1}}{a} = \frac{β - y_{1}}{b} = \frac{γ - z_{1}}{c} = - (\frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{a^{2} + b^{2} + c^{2}}) \dots$
In the given question, $x_{1} = 7, y_{1} = 14, z_{1} = 5, a = 2, b = 4, c = - 1, d = - 2$ By putting these values in (1), we get
$\frac{α - 7}{2} = \frac{β - 14}{4} = \frac{γ - 5}{- 1} = - \frac{63}{21}$
$\Rightarrow α = 1, β = 2 and γ = 8$ Hence, foot of $⊥$ from point to the plane $(1, 2, 8)$

BITSAT-2012

Three Dimensional Geometry

121421 The equation of the plane in normal form passing through the point $A (\overset{―}{a})$ , parallel to a vector $\overset{―}{b}$ and containing $\overset{―}{a}$ vector $\overset{―}{c}$ is

1 r.

\frac{c \times a}{| c \times a |} = | \frac{a \times b}{a \times 6} |

2 r.

\frac{a \times b}{| a \times b |} = \frac{[abc]}{| b \times c |}

3 r.

\frac{b \times c}{| b \times c |} = \frac{[abc]}{| b \times c |}

4 r.[abc]a

= \frac{| b \times c |}{| a \times c |}

Explanation:

C $∴$ Vector normal to plane $n = b \times c$
original image
$n = \frac{b \times c}{| b \times c |}$
Now Ar.n $= 0$
$(r - a) \cdot \frac{(b \times c)}{| b \times c |} = 0 \Rightarrow r \cdot \frac{b \times c}{| b \times c |} = a \cdot \frac{(b \times c)}{| b \times c |}$
$r \cdot \frac{(b \times c)}{| b \times c |} = \frac{[abc]}{| b \times c |}$

TS EAMCET-11.09.2020

Three Dimensional Geometry

121444 Distance between two parallel planes $2 x + y +$ $2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is

1

\frac{5}{2}

2

\frac{7}{2}

3

\frac{9}{2}

4

\frac{3}{2}

Explanation:

B The planes are
$4 x + 2 y + 4 z = 16, 4 x + 2 y + 4 z = - 5$
$\Rightarrow Distance between planes$
$= \frac{16 - (- 5)}{\sqrt{4^{2} + 2^{2} + 4^{2}}} = \frac{21}{6} = \frac{7}{2}$

COMEDK-2013

Three Dimensional Geometry

121410 From a point $P (a, b, c)$ perpendicular $P A, P B$ and drawn to $y z$ and $zx$ planes. Find the equation of the plane $O A B$ , where $O$ is the origin.

1

b c x + c a y + a b z = 0

2

b c x + c a y - a b z = 0

3

b c x - c a y + a b z = 0

4

- bcx + cay + abz = 0

Explanation:

B P(a, b, c) and PA and PB are perpendicular to $YZ$ and $ZX$ planes. Hence, co-ordinate of $A$ and $B$ are $(0, b, c)$ and $(a, 0, c)$ respectively.
Equation of plane passing through $(0, 0, 0)$ ,
$(0, b, c)$ and $(a, 0, c)$ is $| \begin{array}{lll} x & y & z \\ 0 & b & c \\ a & 0 & c \end{array} | = 0$
$\Rightarrow x (b c - 0) - y (0 - a c) + z (0 - a b) = 0$
$\Rightarrow b c x + a c y - a b z = 0$

WB JEE-2021

Three Dimensional Geometry

121404 Equation of line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the plane $x - 2 y - z + 5 = 0$ and $x + y + 3 z = 6$ is

1

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

2

\frac{x - 2}{4} = \frac{y - 3}{3} = \frac{z - 1}{2}

3

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

4

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

Explanation:

A Given
$Point (2, 3, 1)$
$P_{1} = x - 2 y - z + 5 = 0, {\vec{n}}_{1} = \hat{i} - 2 \hat{j} - \hat{k}$
$P_{2} = x + y + 3 z - 6 = 0, {\vec{n}}_{2} = \hat{i} + \hat{j} + 3 \hat{k}$
Intersection of the plane-
$Line, L = {\vec{n}}_{1} \times {\vec{n}}_{2}$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & - 1 \\ 1 & 1 & 3 \end{array} |$
$= \hat{i} (- 6 + 1) - \hat{j} (3 + 1) + \hat{k} (1 + 2) = - 5 \hat{i} - 4 \hat{j} + 3 \hat{k}$
Direction ratios of line $(- 5, - 4, 3)$ i.e. $(a, b, c)$
$∴$ Points $(2, 3, 1)$ i.e. $(x_{1}, y_{1}, z_{1})$
Equation of line, $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$
$= \frac{x - 2}{- 5} = \frac{y - 3}{- 4} = \frac{z - 1}{3}$

Karnataka CET-2015

Three Dimensional Geometry

121419 The foot of the perpendicular from the point (7, $14, 5)$ to the plane $2 x + 4 y - z = 2$ are

1

(1, 2, 8)

2

(3, 2, 8)

3

(5, 10, 6)

4

(9, 18, 4)

Explanation:

A We know that the length of the perpendicular from the point $(x_{1}, y_{1}, z_{1})$ to the plane $ax + by + cz + d = 0$ is $\frac{| {ax}_{1} + {by}_{1} + {cz}_{1} + d |}{\sqrt{a^{2} + b^{2} + c^{2}}}$ and the co-ordinate $(α, β, γ)$ or the foot of the $⊥$ are given by-
$\frac{α - x_{1}}{a} = \frac{β - y_{1}}{b} = \frac{γ - z_{1}}{c} = - (\frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{a^{2} + b^{2} + c^{2}}) \dots$
In the given question, $x_{1} = 7, y_{1} = 14, z_{1} = 5, a = 2, b = 4, c = - 1, d = - 2$ By putting these values in (1), we get
$\frac{α - 7}{2} = \frac{β - 14}{4} = \frac{γ - 5}{- 1} = - \frac{63}{21}$
$\Rightarrow α = 1, β = 2 and γ = 8$ Hence, foot of $⊥$ from point to the plane $(1, 2, 8)$

BITSAT-2012

Three Dimensional Geometry

121421 The equation of the plane in normal form passing through the point $A (\overset{―}{a})$ , parallel to a vector $\overset{―}{b}$ and containing $\overset{―}{a}$ vector $\overset{―}{c}$ is

1 r.

\frac{c \times a}{| c \times a |} = | \frac{a \times b}{a \times 6} |

2 r.

\frac{a \times b}{| a \times b |} = \frac{[abc]}{| b \times c |}

3 r.

\frac{b \times c}{| b \times c |} = \frac{[abc]}{| b \times c |}

4 r.[abc]a

= \frac{| b \times c |}{| a \times c |}

Explanation:

C $∴$ Vector normal to plane $n = b \times c$
original image
$n = \frac{b \times c}{| b \times c |}$
Now Ar.n $= 0$
$(r - a) \cdot \frac{(b \times c)}{| b \times c |} = 0 \Rightarrow r \cdot \frac{b \times c}{| b \times c |} = a \cdot \frac{(b \times c)}{| b \times c |}$
$r \cdot \frac{(b \times c)}{| b \times c |} = \frac{[abc]}{| b \times c |}$

TS EAMCET-11.09.2020

Three Dimensional Geometry

121444 Distance between two parallel planes $2 x + y +$ $2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is

1

\frac{5}{2}

2

\frac{7}{2}

3

\frac{9}{2}

4

\frac{3}{2}

Explanation:

B The planes are
$4 x + 2 y + 4 z = 16, 4 x + 2 y + 4 z = - 5$
$\Rightarrow Distance between planes$
$= \frac{16 - (- 5)}{\sqrt{4^{2} + 2^{2} + 4^{2}}} = \frac{21}{6} = \frac{7}{2}$

COMEDK-2013

Three Dimensional Geometry

121410 From a point $P (a, b, c)$ perpendicular $P A, P B$ and drawn to $y z$ and $zx$ planes. Find the equation of the plane $O A B$ , where $O$ is the origin.

1

b c x + c a y + a b z = 0

2

b c x + c a y - a b z = 0

3

b c x - c a y + a b z = 0

4

- bcx + cay + abz = 0

Explanation:

B P(a, b, c) and PA and PB are perpendicular to $YZ$ and $ZX$ planes. Hence, co-ordinate of $A$ and $B$ are $(0, b, c)$ and $(a, 0, c)$ respectively.
Equation of plane passing through $(0, 0, 0)$ ,
$(0, b, c)$ and $(a, 0, c)$ is $| \begin{array}{lll} x & y & z \\ 0 & b & c \\ a & 0 & c \end{array} | = 0$
$\Rightarrow x (b c - 0) - y (0 - a c) + z (0 - a b) = 0$
$\Rightarrow b c x + a c y - a b z = 0$

WB JEE-2021

Three Dimensional Geometry

121404 Equation of line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the plane $x - 2 y - z + 5 = 0$ and $x + y + 3 z = 6$ is

1

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

2

\frac{x - 2}{4} = \frac{y - 3}{3} = \frac{z - 1}{2}

3

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

4

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

Explanation:

A Given
$Point (2, 3, 1)$
$P_{1} = x - 2 y - z + 5 = 0, {\vec{n}}_{1} = \hat{i} - 2 \hat{j} - \hat{k}$
$P_{2} = x + y + 3 z - 6 = 0, {\vec{n}}_{2} = \hat{i} + \hat{j} + 3 \hat{k}$
Intersection of the plane-
$Line, L = {\vec{n}}_{1} \times {\vec{n}}_{2}$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & - 1 \\ 1 & 1 & 3 \end{array} |$
$= \hat{i} (- 6 + 1) - \hat{j} (3 + 1) + \hat{k} (1 + 2) = - 5 \hat{i} - 4 \hat{j} + 3 \hat{k}$
Direction ratios of line $(- 5, - 4, 3)$ i.e. $(a, b, c)$
$∴$ Points $(2, 3, 1)$ i.e. $(x_{1}, y_{1}, z_{1})$
Equation of line, $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$
$= \frac{x - 2}{- 5} = \frac{y - 3}{- 4} = \frac{z - 1}{3}$

Karnataka CET-2015

Three Dimensional Geometry

121419 The foot of the perpendicular from the point (7, $14, 5)$ to the plane $2 x + 4 y - z = 2$ are

1

(1, 2, 8)

2

(3, 2, 8)

3

(5, 10, 6)

4

(9, 18, 4)

Explanation:

A We know that the length of the perpendicular from the point $(x_{1}, y_{1}, z_{1})$ to the plane $ax + by + cz + d = 0$ is $\frac{| {ax}_{1} + {by}_{1} + {cz}_{1} + d |}{\sqrt{a^{2} + b^{2} + c^{2}}}$ and the co-ordinate $(α, β, γ)$ or the foot of the $⊥$ are given by-
$\frac{α - x_{1}}{a} = \frac{β - y_{1}}{b} = \frac{γ - z_{1}}{c} = - (\frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{a^{2} + b^{2} + c^{2}}) \dots$
In the given question, $x_{1} = 7, y_{1} = 14, z_{1} = 5, a = 2, b = 4, c = - 1, d = - 2$ By putting these values in (1), we get
$\frac{α - 7}{2} = \frac{β - 14}{4} = \frac{γ - 5}{- 1} = - \frac{63}{21}$
$\Rightarrow α = 1, β = 2 and γ = 8$ Hence, foot of $⊥$ from point to the plane $(1, 2, 8)$

BITSAT-2012

Three Dimensional Geometry

121421 The equation of the plane in normal form passing through the point $A (\overset{―}{a})$ , parallel to a vector $\overset{―}{b}$ and containing $\overset{―}{a}$ vector $\overset{―}{c}$ is

1 r.

\frac{c \times a}{| c \times a |} = | \frac{a \times b}{a \times 6} |

2 r.

\frac{a \times b}{| a \times b |} = \frac{[abc]}{| b \times c |}

3 r.

\frac{b \times c}{| b \times c |} = \frac{[abc]}{| b \times c |}

4 r.[abc]a

= \frac{| b \times c |}{| a \times c |}

Explanation:

C $∴$ Vector normal to plane $n = b \times c$
original image
$n = \frac{b \times c}{| b \times c |}$
Now Ar.n $= 0$
$(r - a) \cdot \frac{(b \times c)}{| b \times c |} = 0 \Rightarrow r \cdot \frac{b \times c}{| b \times c |} = a \cdot \frac{(b \times c)}{| b \times c |}$
$r \cdot \frac{(b \times c)}{| b \times c |} = \frac{[abc]}{| b \times c |}$

TS EAMCET-11.09.2020

Three Dimensional Geometry

121444 Distance between two parallel planes $2 x + y +$ $2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is

1

\frac{5}{2}

2

\frac{7}{2}

3

\frac{9}{2}

4

\frac{3}{2}

Explanation:

B The planes are
$4 x + 2 y + 4 z = 16, 4 x + 2 y + 4 z = - 5$
$\Rightarrow Distance between planes$
$= \frac{16 - (- 5)}{\sqrt{4^{2} + 2^{2} + 4^{2}}} = \frac{21}{6} = \frac{7}{2}$

COMEDK-2013

Three Dimensional Geometry

121410 From a point $P (a, b, c)$ perpendicular $P A, P B$ and drawn to $y z$ and $zx$ planes. Find the equation of the plane $O A B$ , where $O$ is the origin.

1

b c x + c a y + a b z = 0

2

b c x + c a y - a b z = 0

3

b c x - c a y + a b z = 0

4

- bcx + cay + abz = 0

Explanation:

B P(a, b, c) and PA and PB are perpendicular to $YZ$ and $ZX$ planes. Hence, co-ordinate of $A$ and $B$ are $(0, b, c)$ and $(a, 0, c)$ respectively.
Equation of plane passing through $(0, 0, 0)$ ,
$(0, b, c)$ and $(a, 0, c)$ is $| \begin{array}{lll} x & y & z \\ 0 & b & c \\ a & 0 & c \end{array} | = 0$
$\Rightarrow x (b c - 0) - y (0 - a c) + z (0 - a b) = 0$
$\Rightarrow b c x + a c y - a b z = 0$

WB JEE-2021

Three Dimensional Geometry

121404 Equation of line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the plane $x - 2 y - z + 5 = 0$ and $x + y + 3 z = 6$ is

1

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

2

\frac{x - 2}{4} = \frac{y - 3}{3} = \frac{z - 1}{2}

3

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

4

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

Explanation:

A Given
$Point (2, 3, 1)$
$P_{1} = x - 2 y - z + 5 = 0, {\vec{n}}_{1} = \hat{i} - 2 \hat{j} - \hat{k}$
$P_{2} = x + y + 3 z - 6 = 0, {\vec{n}}_{2} = \hat{i} + \hat{j} + 3 \hat{k}$
Intersection of the plane-
$Line, L = {\vec{n}}_{1} \times {\vec{n}}_{2}$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & - 1 \\ 1 & 1 & 3 \end{array} |$
$= \hat{i} (- 6 + 1) - \hat{j} (3 + 1) + \hat{k} (1 + 2) = - 5 \hat{i} - 4 \hat{j} + 3 \hat{k}$
Direction ratios of line $(- 5, - 4, 3)$ i.e. $(a, b, c)$
$∴$ Points $(2, 3, 1)$ i.e. $(x_{1}, y_{1}, z_{1})$
Equation of line, $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$
$= \frac{x - 2}{- 5} = \frac{y - 3}{- 4} = \frac{z - 1}{3}$

Karnataka CET-2015

Three Dimensional Geometry

121419 The foot of the perpendicular from the point (7, $14, 5)$ to the plane $2 x + 4 y - z = 2$ are

1

(1, 2, 8)

2

(3, 2, 8)

3

(5, 10, 6)

4

(9, 18, 4)

Explanation:

A We know that the length of the perpendicular from the point $(x_{1}, y_{1}, z_{1})$ to the plane $ax + by + cz + d = 0$ is $\frac{| {ax}_{1} + {by}_{1} + {cz}_{1} + d |}{\sqrt{a^{2} + b^{2} + c^{2}}}$ and the co-ordinate $(α, β, γ)$ or the foot of the $⊥$ are given by-
$\frac{α - x_{1}}{a} = \frac{β - y_{1}}{b} = \frac{γ - z_{1}}{c} = - (\frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{a^{2} + b^{2} + c^{2}}) \dots$
In the given question, $x_{1} = 7, y_{1} = 14, z_{1} = 5, a = 2, b = 4, c = - 1, d = - 2$ By putting these values in (1), we get
$\frac{α - 7}{2} = \frac{β - 14}{4} = \frac{γ - 5}{- 1} = - \frac{63}{21}$
$\Rightarrow α = 1, β = 2 and γ = 8$ Hence, foot of $⊥$ from point to the plane $(1, 2, 8)$

BITSAT-2012

Three Dimensional Geometry

121421 The equation of the plane in normal form passing through the point $A (\overset{―}{a})$ , parallel to a vector $\overset{―}{b}$ and containing $\overset{―}{a}$ vector $\overset{―}{c}$ is

1 r.

\frac{c \times a}{| c \times a |} = | \frac{a \times b}{a \times 6} |

2 r.

\frac{a \times b}{| a \times b |} = \frac{[abc]}{| b \times c |}

3 r.

\frac{b \times c}{| b \times c |} = \frac{[abc]}{| b \times c |}

4 r.[abc]a

= \frac{| b \times c |}{| a \times c |}

Explanation:

C $∴$ Vector normal to plane $n = b \times c$
original image
$n = \frac{b \times c}{| b \times c |}$
Now Ar.n $= 0$
$(r - a) \cdot \frac{(b \times c)}{| b \times c |} = 0 \Rightarrow r \cdot \frac{b \times c}{| b \times c |} = a \cdot \frac{(b \times c)}{| b \times c |}$
$r \cdot \frac{(b \times c)}{| b \times c |} = \frac{[abc]}{| b \times c |}$

TS EAMCET-11.09.2020

Three Dimensional Geometry

121444 Distance between two parallel planes $2 x + y +$ $2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is

1

\frac{5}{2}

2

\frac{7}{2}

3

\frac{9}{2}

4

\frac{3}{2}

Explanation:

B The planes are
$4 x + 2 y + 4 z = 16, 4 x + 2 y + 4 z = - 5$
$\Rightarrow Distance between planes$
$= \frac{16 - (- 5)}{\sqrt{4^{2} + 2^{2} + 4^{2}}} = \frac{21}{6} = \frac{7}{2}$

COMEDK-2013

Three Dimensional Geometry

121410 From a point $P (a, b, c)$ perpendicular $P A, P B$ and drawn to $y z$ and $zx$ planes. Find the equation of the plane $O A B$ , where $O$ is the origin.

1

b c x + c a y + a b z = 0

2

b c x + c a y - a b z = 0

3

b c x - c a y + a b z = 0

4

- bcx + cay + abz = 0

Explanation:

B P(a, b, c) and PA and PB are perpendicular to $YZ$ and $ZX$ planes. Hence, co-ordinate of $A$ and $B$ are $(0, b, c)$ and $(a, 0, c)$ respectively.
Equation of plane passing through $(0, 0, 0)$ ,
$(0, b, c)$ and $(a, 0, c)$ is $| \begin{array}{lll} x & y & z \\ 0 & b & c \\ a & 0 & c \end{array} | = 0$
$\Rightarrow x (b c - 0) - y (0 - a c) + z (0 - a b) = 0$
$\Rightarrow b c x + a c y - a b z = 0$

WB JEE-2021

Three Dimensional Geometry

121404 Equation of line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the plane $x - 2 y - z + 5 = 0$ and $x + y + 3 z = 6$ is

1

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

2

\frac{x - 2}{4} = \frac{y - 3}{3} = \frac{z - 1}{2}

3

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

4

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

Explanation:

A Given
$Point (2, 3, 1)$
$P_{1} = x - 2 y - z + 5 = 0, {\vec{n}}_{1} = \hat{i} - 2 \hat{j} - \hat{k}$
$P_{2} = x + y + 3 z - 6 = 0, {\vec{n}}_{2} = \hat{i} + \hat{j} + 3 \hat{k}$
Intersection of the plane-
$Line, L = {\vec{n}}_{1} \times {\vec{n}}_{2}$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & - 1 \\ 1 & 1 & 3 \end{array} |$
$= \hat{i} (- 6 + 1) - \hat{j} (3 + 1) + \hat{k} (1 + 2) = - 5 \hat{i} - 4 \hat{j} + 3 \hat{k}$
Direction ratios of line $(- 5, - 4, 3)$ i.e. $(a, b, c)$
$∴$ Points $(2, 3, 1)$ i.e. $(x_{1}, y_{1}, z_{1})$
Equation of line, $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$
$= \frac{x - 2}{- 5} = \frac{y - 3}{- 4} = \frac{z - 1}{3}$

Karnataka CET-2015

Three Dimensional Geometry

121419 The foot of the perpendicular from the point (7, $14, 5)$ to the plane $2 x + 4 y - z = 2$ are

1

(1, 2, 8)

2

(3, 2, 8)

3

(5, 10, 6)

4

(9, 18, 4)

Explanation:

A We know that the length of the perpendicular from the point $(x_{1}, y_{1}, z_{1})$ to the plane $ax + by + cz + d = 0$ is $\frac{| {ax}_{1} + {by}_{1} + {cz}_{1} + d |}{\sqrt{a^{2} + b^{2} + c^{2}}}$ and the co-ordinate $(α, β, γ)$ or the foot of the $⊥$ are given by-
$\frac{α - x_{1}}{a} = \frac{β - y_{1}}{b} = \frac{γ - z_{1}}{c} = - (\frac{{ax}_{1} + {by}_{1} + {cz}_{1} + d}{a^{2} + b^{2} + c^{2}}) \dots$
In the given question, $x_{1} = 7, y_{1} = 14, z_{1} = 5, a = 2, b = 4, c = - 1, d = - 2$ By putting these values in (1), we get
$\frac{α - 7}{2} = \frac{β - 14}{4} = \frac{γ - 5}{- 1} = - \frac{63}{21}$
$\Rightarrow α = 1, β = 2 and γ = 8$ Hence, foot of $⊥$ from point to the plane $(1, 2, 8)$

BITSAT-2012

Three Dimensional Geometry

121421 The equation of the plane in normal form passing through the point $A (\overset{―}{a})$ , parallel to a vector $\overset{―}{b}$ and containing $\overset{―}{a}$ vector $\overset{―}{c}$ is

1 r.

\frac{c \times a}{| c \times a |} = | \frac{a \times b}{a \times 6} |

2 r.

\frac{a \times b}{| a \times b |} = \frac{[abc]}{| b \times c |}

3 r.

\frac{b \times c}{| b \times c |} = \frac{[abc]}{| b \times c |}

4 r.[abc]a

= \frac{| b \times c |}{| a \times c |}

Explanation:

C $∴$ Vector normal to plane $n = b \times c$
original image
$n = \frac{b \times c}{| b \times c |}$
Now Ar.n $= 0$
$(r - a) \cdot \frac{(b \times c)}{| b \times c |} = 0 \Rightarrow r \cdot \frac{b \times c}{| b \times c |} = a \cdot \frac{(b \times c)}{| b \times c |}$
$r \cdot \frac{(b \times c)}{| b \times c |} = \frac{[abc]}{| b \times c |}$

TS EAMCET-11.09.2020