QBManager

Three Dimensional Geometry

121200 The equation of the line joining the points $(- 3, 4, 11)$ and $(1, - 2, 7)$ is

1

\frac{x + 3}{2} = \frac{y - 4}{3} = \frac{z - 11}{4}

2

\frac{x + 3}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}

3

\frac{x + 3}{- 2} = \frac{y + 4}{3} = \frac{z + 11}{4}

4

\frac{x + 3}{2} = \frac{y + 4}{- 3} = \frac{z + 11}{2}

Explanation:

B Given,
Point, $(- 3, 4, 11)$ and $(1, - 2, 7)$
$∵$ Direction ratios of the given, point-
$[1 - (- 3), (- 2 - 4), (7, - 11)]$
$(4, - 6, - 4)$ or $(- 2, 3, 2)$
i.e. $(4, - 6, - 4)$ or $(- 2, 3, 2)$
$∴$ Equation of line, $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$
or $\frac{x - (- 3)}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}$
$\Rightarrow \frac{x + 3}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}$

Karnataka CET-2021

Three Dimensional Geometry

121201 If the planes $\vec{r} \cdot (2 \hat{i} - λ \hat{j} + \hat{k}) = 3$ and $\vec{r} \cdot (4 \hat{i} + \hat{j} - μ \hat{k}) = 5$ are parallel, then what is $λ$ equal to?

1

1 / 2

2 1

3

- 1 / 2

4 -1

Explanation:

C Given,
Equation of planes :
$\vec{r} (2 \hat{i} - λ \hat{j} + \hat{k}) = 3$
$\vec{r} (4 \hat{i} + \hat{j} - μ \hat{k}) = 5$
and
Both planes are parallel then,
$\frac{2}{4} = \frac{- λ}{1} = \frac{1}{μ}$
$λ = \frac{- 1}{2}$

SCRA-2012

Three Dimensional Geometry

121203 The Cartesian equation of the line passing through the point $(- 1, 3, - 2)$ and perpendicular to the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x + 2}{- 3} = \frac{y - 1}{2} = \frac{z + 1}{5}$ is

1

\frac{x - 1}{2} = \frac{y + 3}{7} = \frac{z - 2}{4}

2

\frac{x - 1}{- 2} = \frac{y + 3}{- 7} = \frac{z - 2}{- 4}

3

\frac{x + 1}{2} = \frac{y + 3}{7} = \frac{z + 2}{4}

4

\frac{x + 1}{2} = \frac{y - 3}{- 7} = \frac{z + 2}{4}

Explanation:

D Given,
point $A (- 1, 3, - 2)$
$1^{st}$ line $: l_{1} : \frac{x}{1} = \frac{y}{2} = \frac{z}{3}$
$l_{2} : \frac{x + 2}{- 3} = \frac{y - 1}{2} = \frac{z + 1}{5}$
Equation of the line passing through the point ( $- 1, 3, -$ 2) and assuming the direction ratios as a, b, c is given by-
$\frac{x + 1}{a} = \frac{y - 3}{b} = \frac{z + 2}{c}$
As the required line is perpendicular to the line $l_{1}$ and $l_{2}$ , $\Rightarrow a + 2 b + 3 c = 0$ and
$- 3 a + 2 b + 5 c = 0$
Solving for $a, b, c$ we get,
$\frac{a}{10 - 6} = \frac{b}{- 9 - 5} = \frac{c}{2 + 6}$
$\Rightarrow \frac{a}{4} = \frac{- b}{14} = \frac{c}{8}$
$\Rightarrow \frac{a}{2} = \frac{- b}{7} = \frac{c}{4}$
From (iii), the cartesian equation of the line is
$\frac{x + 1}{2} = \frac{y - 3}{- 7} = \frac{z + 2}{4}$

AP EAMCET-22.09.2020

Three Dimensional Geometry

121204 The equation of the plane, which bisects the line joining the points $(1, 2, 3)$ and $(3, 4, 5)$ at right angles is

1

x + y + z = 0

2

x + y - z = 0

3

x + y + z = 9

4

x + y - z + 9 = 0

Explanation:

C The given points are $A (1, 2, 3)$ and $B (3, 4, 5)$ The direction ratios of line segment $A B$ is given by
$(x_{2} - x_{1}), (y_{2} - y_{1}), (z_{2} - z_{1})$
i.e. $(3 - 1), (4 - 2), (5 - 3) = (2, 2, 2) = (1, 1, 1)$
Since, the plane bisects $A B$ at right angles. $A B$ is the normal to the plane. Which is $n$
Therefore, $n = \hat{i} + \hat{j} + \hat{k}$
Let $c$ be the mid-point of $AB$
$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}, \frac{z_{1} + z_{2}}{2})$
$c = (\frac{1 + 3}{2}, \frac{2 + 4}{2}, \frac{3 + 5}{2}), c = (2, 3, 4)$
Let this vector be $a = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
Hence, the required equation of vector
${r - (2 \hat{i} + 3 \hat{j} + 4 \hat{k})} (\hat{i} + \hat{j} + \hat{k}) = 0$
We know that, $r = x \hat{i} + y \hat{j} + z \hat{k}$
${(x \hat{i} + y \hat{j} + z \hat{k}) - (2 \hat{i} + 3 \hat{j} + 4 \hat{k})} (\hat{i} + \hat{j} + \hat{k}) = 0$
$(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})$
$\Rightarrow x + y + z = 2 + 3 + 4 = 9$

WB JEE-2019

Three Dimensional Geometry

121200 The equation of the line joining the points $(- 3, 4, 11)$ and $(1, - 2, 7)$ is

1

\frac{x + 3}{2} = \frac{y - 4}{3} = \frac{z - 11}{4}

2

\frac{x + 3}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}

3

\frac{x + 3}{- 2} = \frac{y + 4}{3} = \frac{z + 11}{4}

4

\frac{x + 3}{2} = \frac{y + 4}{- 3} = \frac{z + 11}{2}

Explanation:

B Given,
Point, $(- 3, 4, 11)$ and $(1, - 2, 7)$
$∵$ Direction ratios of the given, point-
$[1 - (- 3), (- 2 - 4), (7, - 11)]$
$(4, - 6, - 4)$ or $(- 2, 3, 2)$
i.e. $(4, - 6, - 4)$ or $(- 2, 3, 2)$
$∴$ Equation of line, $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$
or $\frac{x - (- 3)}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}$
$\Rightarrow \frac{x + 3}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}$

Karnataka CET-2021

Three Dimensional Geometry

121201 If the planes $\vec{r} \cdot (2 \hat{i} - λ \hat{j} + \hat{k}) = 3$ and $\vec{r} \cdot (4 \hat{i} + \hat{j} - μ \hat{k}) = 5$ are parallel, then what is $λ$ equal to?

1

1 / 2

2 1

3

- 1 / 2

4 -1

Explanation:

C Given,
Equation of planes :
$\vec{r} (2 \hat{i} - λ \hat{j} + \hat{k}) = 3$
$\vec{r} (4 \hat{i} + \hat{j} - μ \hat{k}) = 5$
and
Both planes are parallel then,
$\frac{2}{4} = \frac{- λ}{1} = \frac{1}{μ}$
$λ = \frac{- 1}{2}$

SCRA-2012

Three Dimensional Geometry

121203 The Cartesian equation of the line passing through the point $(- 1, 3, - 2)$ and perpendicular to the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x + 2}{- 3} = \frac{y - 1}{2} = \frac{z + 1}{5}$ is

1

\frac{x - 1}{2} = \frac{y + 3}{7} = \frac{z - 2}{4}

2

\frac{x - 1}{- 2} = \frac{y + 3}{- 7} = \frac{z - 2}{- 4}

3

\frac{x + 1}{2} = \frac{y + 3}{7} = \frac{z + 2}{4}

4

\frac{x + 1}{2} = \frac{y - 3}{- 7} = \frac{z + 2}{4}

Explanation:

D Given,
point $A (- 1, 3, - 2)$
$1^{st}$ line $: l_{1} : \frac{x}{1} = \frac{y}{2} = \frac{z}{3}$
$l_{2} : \frac{x + 2}{- 3} = \frac{y - 1}{2} = \frac{z + 1}{5}$
Equation of the line passing through the point ( $- 1, 3, -$ 2) and assuming the direction ratios as a, b, c is given by-
$\frac{x + 1}{a} = \frac{y - 3}{b} = \frac{z + 2}{c}$
As the required line is perpendicular to the line $l_{1}$ and $l_{2}$ , $\Rightarrow a + 2 b + 3 c = 0$ and
$- 3 a + 2 b + 5 c = 0$
Solving for $a, b, c$ we get,
$\frac{a}{10 - 6} = \frac{b}{- 9 - 5} = \frac{c}{2 + 6}$
$\Rightarrow \frac{a}{4} = \frac{- b}{14} = \frac{c}{8}$
$\Rightarrow \frac{a}{2} = \frac{- b}{7} = \frac{c}{4}$
From (iii), the cartesian equation of the line is
$\frac{x + 1}{2} = \frac{y - 3}{- 7} = \frac{z + 2}{4}$

AP EAMCET-22.09.2020

Three Dimensional Geometry

121204 The equation of the plane, which bisects the line joining the points $(1, 2, 3)$ and $(3, 4, 5)$ at right angles is

1

x + y + z = 0

2

x + y - z = 0

3

x + y + z = 9

4

x + y - z + 9 = 0

Explanation:

C The given points are $A (1, 2, 3)$ and $B (3, 4, 5)$ The direction ratios of line segment $A B$ is given by
$(x_{2} - x_{1}), (y_{2} - y_{1}), (z_{2} - z_{1})$
i.e. $(3 - 1), (4 - 2), (5 - 3) = (2, 2, 2) = (1, 1, 1)$
Since, the plane bisects $A B$ at right angles. $A B$ is the normal to the plane. Which is $n$
Therefore, $n = \hat{i} + \hat{j} + \hat{k}$
Let $c$ be the mid-point of $AB$
$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}, \frac{z_{1} + z_{2}}{2})$
$c = (\frac{1 + 3}{2}, \frac{2 + 4}{2}, \frac{3 + 5}{2}), c = (2, 3, 4)$
Let this vector be $a = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
Hence, the required equation of vector
${r - (2 \hat{i} + 3 \hat{j} + 4 \hat{k})} (\hat{i} + \hat{j} + \hat{k}) = 0$
We know that, $r = x \hat{i} + y \hat{j} + z \hat{k}$
${(x \hat{i} + y \hat{j} + z \hat{k}) - (2 \hat{i} + 3 \hat{j} + 4 \hat{k})} (\hat{i} + \hat{j} + \hat{k}) = 0$
$(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})$
$\Rightarrow x + y + z = 2 + 3 + 4 = 9$

WB JEE-2019

Three Dimensional Geometry

121200 The equation of the line joining the points $(- 3, 4, 11)$ and $(1, - 2, 7)$ is

1

\frac{x + 3}{2} = \frac{y - 4}{3} = \frac{z - 11}{4}

2

\frac{x + 3}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}

3

\frac{x + 3}{- 2} = \frac{y + 4}{3} = \frac{z + 11}{4}

4

\frac{x + 3}{2} = \frac{y + 4}{- 3} = \frac{z + 11}{2}

Explanation:

B Given,
Point, $(- 3, 4, 11)$ and $(1, - 2, 7)$
$∵$ Direction ratios of the given, point-
$[1 - (- 3), (- 2 - 4), (7, - 11)]$
$(4, - 6, - 4)$ or $(- 2, 3, 2)$
i.e. $(4, - 6, - 4)$ or $(- 2, 3, 2)$
$∴$ Equation of line, $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$
or $\frac{x - (- 3)}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}$
$\Rightarrow \frac{x + 3}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}$

Karnataka CET-2021

Three Dimensional Geometry

121201 If the planes $\vec{r} \cdot (2 \hat{i} - λ \hat{j} + \hat{k}) = 3$ and $\vec{r} \cdot (4 \hat{i} + \hat{j} - μ \hat{k}) = 5$ are parallel, then what is $λ$ equal to?

1

1 / 2

2 1

3

- 1 / 2

4 -1

Explanation:

C Given,
Equation of planes :
$\vec{r} (2 \hat{i} - λ \hat{j} + \hat{k}) = 3$
$\vec{r} (4 \hat{i} + \hat{j} - μ \hat{k}) = 5$
and
Both planes are parallel then,
$\frac{2}{4} = \frac{- λ}{1} = \frac{1}{μ}$
$λ = \frac{- 1}{2}$

SCRA-2012

Three Dimensional Geometry

121203 The Cartesian equation of the line passing through the point $(- 1, 3, - 2)$ and perpendicular to the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x + 2}{- 3} = \frac{y - 1}{2} = \frac{z + 1}{5}$ is

1

\frac{x - 1}{2} = \frac{y + 3}{7} = \frac{z - 2}{4}

2

\frac{x - 1}{- 2} = \frac{y + 3}{- 7} = \frac{z - 2}{- 4}

3

\frac{x + 1}{2} = \frac{y + 3}{7} = \frac{z + 2}{4}

4

\frac{x + 1}{2} = \frac{y - 3}{- 7} = \frac{z + 2}{4}

Explanation:

D Given,
point $A (- 1, 3, - 2)$
$1^{st}$ line $: l_{1} : \frac{x}{1} = \frac{y}{2} = \frac{z}{3}$
$l_{2} : \frac{x + 2}{- 3} = \frac{y - 1}{2} = \frac{z + 1}{5}$
Equation of the line passing through the point ( $- 1, 3, -$ 2) and assuming the direction ratios as a, b, c is given by-
$\frac{x + 1}{a} = \frac{y - 3}{b} = \frac{z + 2}{c}$
As the required line is perpendicular to the line $l_{1}$ and $l_{2}$ , $\Rightarrow a + 2 b + 3 c = 0$ and
$- 3 a + 2 b + 5 c = 0$
Solving for $a, b, c$ we get,
$\frac{a}{10 - 6} = \frac{b}{- 9 - 5} = \frac{c}{2 + 6}$
$\Rightarrow \frac{a}{4} = \frac{- b}{14} = \frac{c}{8}$
$\Rightarrow \frac{a}{2} = \frac{- b}{7} = \frac{c}{4}$
From (iii), the cartesian equation of the line is
$\frac{x + 1}{2} = \frac{y - 3}{- 7} = \frac{z + 2}{4}$

AP EAMCET-22.09.2020

Three Dimensional Geometry

121204 The equation of the plane, which bisects the line joining the points $(1, 2, 3)$ and $(3, 4, 5)$ at right angles is

1

x + y + z = 0

2

x + y - z = 0

3

x + y + z = 9

4

x + y - z + 9 = 0

Explanation:

C The given points are $A (1, 2, 3)$ and $B (3, 4, 5)$ The direction ratios of line segment $A B$ is given by
$(x_{2} - x_{1}), (y_{2} - y_{1}), (z_{2} - z_{1})$
i.e. $(3 - 1), (4 - 2), (5 - 3) = (2, 2, 2) = (1, 1, 1)$
Since, the plane bisects $A B$ at right angles. $A B$ is the normal to the plane. Which is $n$
Therefore, $n = \hat{i} + \hat{j} + \hat{k}$
Let $c$ be the mid-point of $AB$
$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}, \frac{z_{1} + z_{2}}{2})$
$c = (\frac{1 + 3}{2}, \frac{2 + 4}{2}, \frac{3 + 5}{2}), c = (2, 3, 4)$
Let this vector be $a = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
Hence, the required equation of vector
${r - (2 \hat{i} + 3 \hat{j} + 4 \hat{k})} (\hat{i} + \hat{j} + \hat{k}) = 0$
We know that, $r = x \hat{i} + y \hat{j} + z \hat{k}$
${(x \hat{i} + y \hat{j} + z \hat{k}) - (2 \hat{i} + 3 \hat{j} + 4 \hat{k})} (\hat{i} + \hat{j} + \hat{k}) = 0$
$(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})$
$\Rightarrow x + y + z = 2 + 3 + 4 = 9$

WB JEE-2019

Three Dimensional Geometry

121200 The equation of the line joining the points $(- 3, 4, 11)$ and $(1, - 2, 7)$ is

1

\frac{x + 3}{2} = \frac{y - 4}{3} = \frac{z - 11}{4}

2

\frac{x + 3}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}

3

\frac{x + 3}{- 2} = \frac{y + 4}{3} = \frac{z + 11}{4}

4

\frac{x + 3}{2} = \frac{y + 4}{- 3} = \frac{z + 11}{2}

Explanation:

B Given,
Point, $(- 3, 4, 11)$ and $(1, - 2, 7)$
$∵$ Direction ratios of the given, point-
$[1 - (- 3), (- 2 - 4), (7, - 11)]$
$(4, - 6, - 4)$ or $(- 2, 3, 2)$
i.e. $(4, - 6, - 4)$ or $(- 2, 3, 2)$
$∴$ Equation of line, $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$
or $\frac{x - (- 3)}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}$
$\Rightarrow \frac{x + 3}{- 2} = \frac{y - 4}{3} = \frac{z - 11}{2}$

Karnataka CET-2021

Three Dimensional Geometry

121201 If the planes $\vec{r} \cdot (2 \hat{i} - λ \hat{j} + \hat{k}) = 3$ and $\vec{r} \cdot (4 \hat{i} + \hat{j} - μ \hat{k}) = 5$ are parallel, then what is $λ$ equal to?

1

1 / 2

2 1

3

- 1 / 2

4 -1

Explanation:

C Given,
Equation of planes :
$\vec{r} (2 \hat{i} - λ \hat{j} + \hat{k}) = 3$
$\vec{r} (4 \hat{i} + \hat{j} - μ \hat{k}) = 5$
and
Both planes are parallel then,
$\frac{2}{4} = \frac{- λ}{1} = \frac{1}{μ}$
$λ = \frac{- 1}{2}$

SCRA-2012

Three Dimensional Geometry

121203 The Cartesian equation of the line passing through the point $(- 1, 3, - 2)$ and perpendicular to the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x + 2}{- 3} = \frac{y - 1}{2} = \frac{z + 1}{5}$ is

1

\frac{x - 1}{2} = \frac{y + 3}{7} = \frac{z - 2}{4}

2

\frac{x - 1}{- 2} = \frac{y + 3}{- 7} = \frac{z - 2}{- 4}

3

\frac{x + 1}{2} = \frac{y + 3}{7} = \frac{z + 2}{4}

4

\frac{x + 1}{2} = \frac{y - 3}{- 7} = \frac{z + 2}{4}

Explanation:

D Given,
point $A (- 1, 3, - 2)$
$1^{st}$ line $: l_{1} : \frac{x}{1} = \frac{y}{2} = \frac{z}{3}$
$l_{2} : \frac{x + 2}{- 3} = \frac{y - 1}{2} = \frac{z + 1}{5}$
Equation of the line passing through the point ( $- 1, 3, -$ 2) and assuming the direction ratios as a, b, c is given by-
$\frac{x + 1}{a} = \frac{y - 3}{b} = \frac{z + 2}{c}$
As the required line is perpendicular to the line $l_{1}$ and $l_{2}$ , $\Rightarrow a + 2 b + 3 c = 0$ and
$- 3 a + 2 b + 5 c = 0$
Solving for $a, b, c$ we get,
$\frac{a}{10 - 6} = \frac{b}{- 9 - 5} = \frac{c}{2 + 6}$
$\Rightarrow \frac{a}{4} = \frac{- b}{14} = \frac{c}{8}$
$\Rightarrow \frac{a}{2} = \frac{- b}{7} = \frac{c}{4}$
From (iii), the cartesian equation of the line is
$\frac{x + 1}{2} = \frac{y - 3}{- 7} = \frac{z + 2}{4}$

AP EAMCET-22.09.2020

Three Dimensional Geometry

121204 The equation of the plane, which bisects the line joining the points $(1, 2, 3)$ and $(3, 4, 5)$ at right angles is

1

x + y + z = 0

2

x + y - z = 0

3

x + y + z = 9

4

x + y - z + 9 = 0

Explanation:

C The given points are $A (1, 2, 3)$ and $B (3, 4, 5)$ The direction ratios of line segment $A B$ is given by
$(x_{2} - x_{1}), (y_{2} - y_{1}), (z_{2} - z_{1})$
i.e. $(3 - 1), (4 - 2), (5 - 3) = (2, 2, 2) = (1, 1, 1)$
Since, the plane bisects $A B$ at right angles. $A B$ is the normal to the plane. Which is $n$
Therefore, $n = \hat{i} + \hat{j} + \hat{k}$
Let $c$ be the mid-point of $AB$
$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}, \frac{z_{1} + z_{2}}{2})$
$c = (\frac{1 + 3}{2}, \frac{2 + 4}{2}, \frac{3 + 5}{2}), c = (2, 3, 4)$
Let this vector be $a = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
Hence, the required equation of vector
${r - (2 \hat{i} + 3 \hat{j} + 4 \hat{k})} (\hat{i} + \hat{j} + \hat{k}) = 0$
We know that, $r = x \hat{i} + y \hat{j} + z \hat{k}$
${(x \hat{i} + y \hat{j} + z \hat{k}) - (2 \hat{i} + 3 \hat{j} + 4 \hat{k})} (\hat{i} + \hat{j} + \hat{k}) = 0$
$(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})$
$\Rightarrow x + y + z = 2 + 3 + 4 = 9$

WB JEE-2019

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: