QBManager

Three Dimensional Geometry

121231 Let the line $\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lies on P ?

1 $(0,-2,-2)$

2 $(-5,0,-1)$

3 $(3,-1,0)$

4 $(0,4,5)$

JEE Main-28.07.2022

Three Dimensional Geometry

121232 The shortest distance between the lines $\frac{x+7}{-6}=\frac{y-6}{7}=z$ and $\frac{7-x}{2}=y-2=z-6$ is

1 $2 \sqrt{29}$

2 1

3 $\sqrt{\frac{37}{29}}$

4 $\sqrt{\frac{29}{2}}$

JEE Main-25.07.2022

Three Dimensional Geometry

121236 If the vectors $\alpha=\hat{\mathbf{i}}+\mathbf{a} \hat{\mathbf{j}}+\mathbf{a}^2 \hat{\mathbf{k}}, \boldsymbol{\beta}=\hat{\mathbf{i}}+\mathbf{b} \hat{\mathbf{j}}+\mathbf{b}^2 \hat{\mathbf{k}}$ and $\gamma=\hat{\mathbf{i}}+\mathbf{c} \hat{\mathbf{j}}+\mathbf{c}^2 \hat{\mathbf{k}}$ are three non- coplanar vectors and $| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{2} \end{array} | = 0$ , then the value of abc is

1 1

2 0

3 -1

4 2

Explanation:

C Given,
$α = \hat{i} + a \hat{j} + a^{2} \hat{k}$
$β = \hat{i} + b \hat{j} + b^{2} \hat{k}$
$γ = \hat{i} + c \hat{j} + c^{2} \hat{k}$
$α, β$ and $γ$ are non-coplanar
So, $abc \neq 0$
Now,
$| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = 0$
We can write as -
$\begin{aligned} | \begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array} | + | \begin{array}{lll} a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3} \end{array} | = 0 \\ \Rightarrow | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | + a b c | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0 \\ \Rightarrow | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | . (1 + abc) = 0 \end{aligned}$
Now, as, $| \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | \neq 0$
Hence,
$\Rightarrow 1 + abc = 0$
$abc = - 1$

WB JEE-2020

Three Dimensional Geometry

121238 The value of ' $x$ ' for which the points $(1, 2, 1)$ . $(0, 1, 3), (1, 0, 1)$ and $(2, 0, x)$ are coplanar is

1

x = 0

2

x = 1

3

x = - 1

4

x = 2

Explanation:

C Given,
$A (1, 2, 1), B (0, 1, 3), C (1, 0, 1)$ and $D (2, 0, x)$ are coplanar
$\vec{AB} = \vec{B} - \vec{A}$
$= (0 \hat{i} + \hat{j} + 3 \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$= - \hat{i} - \hat{j} + 2 \hat{k}$
$\vec{AC} = \vec{C} - \vec{A}$
$= (\hat{i} + 0 \hat{j} + \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$b_{1} \times b_{2} = 12 \hat{i} + 8 \hat{j} - 8 \hat{k}$
$\vec{AD} = \vec{D} - \vec{A}$
$= (2 \hat{i} + 0 \hat{j} + x \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$= \hat{i} - 2 \hat{j} + (x - 1) \hat{k}$
Since, A, B, C and D are coplanar, then
$[\begin{array}{lll} \vec{AB} & \vec{AC} & \vec{AD} \end{array}] = 0$
$[\begin{array}{ccc} - 1 & - 1 & 2 \\ 0 & - 2 & 0 \\ 1 & - 2 & x - 1 \end{array}] = 0$
$\Rightarrow - 1 [- 2 (x - 1)] + 1 (0) + 2 (+ 2) = 0$
$\Rightarrow - (- 2 x + 2) + 0 + 4 = 0$
$\Rightarrow 2 x - 2 + 4 = 0$
$2 x + 2 = 0$ Hence, $x = - 1$

J and K CET-2017

Three Dimensional Geometry

121212 Lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- K}$ and $\frac{x - 1}{K} = \frac{y - 4}{2} =$ $\frac{z - 5}{1}$ are coplanar if

1

K = 2

2

K = 0

3

K = 3

4

K = - 1

Explanation:

B Given,
$\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}} is compared$
with $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- K}$
Then, $x_{1} = 2, y_{1} = 3, z_{1} = 4$
Direction ratios, ( $a_{1} = 1, b_{1} = 1, c_{1} = - K)$
$∴$ And, $\frac{x - 1}{K} = \frac{x - 4}{2} = \frac{z - 5}{1}$
$x_{2} = 1, y_{2} = 4, z_{2} = 5$
Direction ratios $(a_{2} = K, b_{2} = 2, c_{2} = 1)$
The two line will be coplanar If,
$\begin{aligned} | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0 \\ \Rightarrow | \begin{array}{ccc} 1 - 2 & 4 - 3 & 5 - 4 \\ 1 & 1 & - K \\ K & 2 & 1 \end{array} | = 0 \end{aligned}$
$\Rightarrow - 1 (1 + 2 K) - 1 (1 + K^{2}) + 1 (2 - K) = 0$
$\Rightarrow - 1 - 2 K - 1 - K^{2} + 2 - K = 0$
$\Rightarrow - 2 K - K^{2} - K = 0$
$\Rightarrow K^{2} + 3 K = 0$
$\Rightarrow K (K + 3) = 0$
$Hence, K = (0, - 3)$ Hence,

BITSAT-2016

Three Dimensional Geometry

121231 Let the line $\frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lies on P ?

1

(0, - 2, - 2)

2

(- 5, 0, - 1)

3

(3, - 1, 0)

4

(0, 4, 5)

Explanation:

D Given,
$l_{1} : \frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$
$∴$ The point $(1, 2, 3)$ and direction ratio $(λ, 1, 2)$
And, $l_{2} : \frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$
$∵$ The point $(- 26, - 18, - 28)$ and direction ratio $(- 2$ , $3, λ)$
$l_{1}$ and $l_{2}$ are coplanar
We know that, the lines are coplanar
$\Rightarrow | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0$
$\Rightarrow | \begin{array}{ccc} - 27 & - 20 & - 31 \\ λ & 1 & 2 \\ - 2 & 3 & λ \end{array} | = 0$
$\Rightarrow λ^{2} + 9 - 6 λ = 0$
$\Rightarrow λ = \pm 3$
Now, vector perpendicular to plane $P$ which contains given lines is -
$| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ - 2 & 3 & 3 \end{array} | = - 3 \hat{i} - 13 \hat{j} + 11 \hat{k}$
Equation of required plane $P$ is,
$- 3 (x - 1) - 13 (y - 2) + 11 (z - 3) = 0$
$\Rightarrow 3 x + 13 y - 11 z + 4 = 0$
$∴ (0, 4, 5)$ does not lie on plane $P$ .

JEE Main-28.07.2022

Three Dimensional Geometry

121232 The shortest distance between the lines $\frac{x + 7}{- 6} = \frac{y - 6}{7} = z$ and $\frac{7 - x}{2} = y - 2 = z - 6$ is

1

2 \sqrt{29}

2 1

3

\sqrt{\frac{37}{29}}

4

\sqrt{\frac{29}{2}}

Explanation:

A Given,
$l_{1} : \frac{x + 7}{- 6} = \frac{y - 6}{7} = \frac{z - 0}{1}$
And, $l_{2} : \frac{x - 7}{- 2} = \frac{y - 2}{1} = \frac{z - 6}{1}$
$(x_{1}, y_{1}, z_{1}) = (- 7, 6, 0)$
$(a_{1}, b_{1}, c_{1}) = (- 6, 7, 1)$
And, $(x_{2}, y_{2}, z_{2}) = (7, 2, 6)$
$(a_{2}, b_{2}, c_{2}) = (- 2, 1, 1)$
Shortest distance between two lines
$\begin{aligned} d = \frac{| \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} |}{\sqrt{{(a_{1} b_{2} - a_{2} b_{1})}^{2} + {(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2}}} \\ d = \frac{| \begin{array}{ccc} 14 & - 4 & 6 \\ - 6 & 7 & 1 \\ - 2 & 1 & 1 \end{array} |}{\sqrt{(- 6 + 14)^{2} + (7 - 1)^{2} + (- 2 + 6)^{2}}} \end{aligned}$
$\Rightarrow d = | \frac{14 (7 - 1) + 4 (- 6 + 2) + 6 (- 6 + 14)}{\sqrt{64 + 36 + 16}} |$
$\Rightarrow d = \frac{116}{\sqrt{116}} = \sqrt{116}$ Hence, $d = 2 \sqrt{29}$

JEE Main-25.07.2022

Three Dimensional Geometry

121236 If the vectors $α = \hat{i} + a \hat{j} + a^{2} \hat{k}, \boldsymbol β = \hat{i} + b \hat{j} + b^{2} \hat{k}$ and $γ = \hat{i} + c \hat{j} + c^{2} \hat{k}$ are three non- coplanar vectors and $| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{2} \end{array} | = 0$ , then the value of abc is

1 1

2 0

3 -1

4 2

Explanation:

C Given,
$α = \hat{i} + a \hat{j} + a^{2} \hat{k}$
$β = \hat{i} + b \hat{j} + b^{2} \hat{k}$
$γ = \hat{i} + c \hat{j} + c^{2} \hat{k}$
$α, β$ and $γ$ are non-coplanar
So, $abc \neq 0$
Now,
$| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = 0$
We can write as -
$\begin{aligned} | \begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array} | + | \begin{array}{lll} a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3} \end{array} | = 0 \\ \Rightarrow | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | + a b c | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0 \\ \Rightarrow | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | . (1 + abc) = 0 \end{aligned}$
Now, as, $| \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | \neq 0$
Hence,
$\Rightarrow 1 + abc = 0$
$abc = - 1$

WB JEE-2020

Three Dimensional Geometry

121238 The value of ' $x$ ' for which the points $(1, 2, 1)$ . $(0, 1, 3), (1, 0, 1)$ and $(2, 0, x)$ are coplanar is

1

x = 0

2

x = 1

3

x = - 1

4

x = 2

Explanation:

C Given,
$A (1, 2, 1), B (0, 1, 3), C (1, 0, 1)$ and $D (2, 0, x)$ are coplanar
$\vec{AB} = \vec{B} - \vec{A}$
$= (0 \hat{i} + \hat{j} + 3 \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$= - \hat{i} - \hat{j} + 2 \hat{k}$
$\vec{AC} = \vec{C} - \vec{A}$
$= (\hat{i} + 0 \hat{j} + \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$b_{1} \times b_{2} = 12 \hat{i} + 8 \hat{j} - 8 \hat{k}$
$\vec{AD} = \vec{D} - \vec{A}$
$= (2 \hat{i} + 0 \hat{j} + x \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$= \hat{i} - 2 \hat{j} + (x - 1) \hat{k}$
Since, A, B, C and D are coplanar, then
$[\begin{array}{lll} \vec{AB} & \vec{AC} & \vec{AD} \end{array}] = 0$
$[\begin{array}{ccc} - 1 & - 1 & 2 \\ 0 & - 2 & 0 \\ 1 & - 2 & x - 1 \end{array}] = 0$
$\Rightarrow - 1 [- 2 (x - 1)] + 1 (0) + 2 (+ 2) = 0$
$\Rightarrow - (- 2 x + 2) + 0 + 4 = 0$
$\Rightarrow 2 x - 2 + 4 = 0$
$2 x + 2 = 0$ Hence, $x = - 1$

J and K CET-2017

Three Dimensional Geometry

121212 Lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- K}$ and $\frac{x - 1}{K} = \frac{y - 4}{2} =$ $\frac{z - 5}{1}$ are coplanar if

1

K = 2

2

K = 0

3

K = 3

4

K = - 1

Explanation:

B Given,
$\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}} is compared$
with $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- K}$
Then, $x_{1} = 2, y_{1} = 3, z_{1} = 4$
Direction ratios, ( $a_{1} = 1, b_{1} = 1, c_{1} = - K)$
$∴$ And, $\frac{x - 1}{K} = \frac{x - 4}{2} = \frac{z - 5}{1}$
$x_{2} = 1, y_{2} = 4, z_{2} = 5$
Direction ratios $(a_{2} = K, b_{2} = 2, c_{2} = 1)$
The two line will be coplanar If,
$\begin{aligned} | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0 \\ \Rightarrow | \begin{array}{ccc} 1 - 2 & 4 - 3 & 5 - 4 \\ 1 & 1 & - K \\ K & 2 & 1 \end{array} | = 0 \end{aligned}$
$\Rightarrow - 1 (1 + 2 K) - 1 (1 + K^{2}) + 1 (2 - K) = 0$
$\Rightarrow - 1 - 2 K - 1 - K^{2} + 2 - K = 0$
$\Rightarrow - 2 K - K^{2} - K = 0$
$\Rightarrow K^{2} + 3 K = 0$
$\Rightarrow K (K + 3) = 0$
$Hence, K = (0, - 3)$ Hence,

BITSAT-2016

Three Dimensional Geometry

121231 Let the line $\frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lies on P ?

1

(0, - 2, - 2)

2

(- 5, 0, - 1)

3

(3, - 1, 0)

4

(0, 4, 5)

Explanation:

D Given,
$l_{1} : \frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$
$∴$ The point $(1, 2, 3)$ and direction ratio $(λ, 1, 2)$
And, $l_{2} : \frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$
$∵$ The point $(- 26, - 18, - 28)$ and direction ratio $(- 2$ , $3, λ)$
$l_{1}$ and $l_{2}$ are coplanar
We know that, the lines are coplanar
$\Rightarrow | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0$
$\Rightarrow | \begin{array}{ccc} - 27 & - 20 & - 31 \\ λ & 1 & 2 \\ - 2 & 3 & λ \end{array} | = 0$
$\Rightarrow λ^{2} + 9 - 6 λ = 0$
$\Rightarrow λ = \pm 3$
Now, vector perpendicular to plane $P$ which contains given lines is -
$| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ - 2 & 3 & 3 \end{array} | = - 3 \hat{i} - 13 \hat{j} + 11 \hat{k}$
Equation of required plane $P$ is,
$- 3 (x - 1) - 13 (y - 2) + 11 (z - 3) = 0$
$\Rightarrow 3 x + 13 y - 11 z + 4 = 0$
$∴ (0, 4, 5)$ does not lie on plane $P$ .

JEE Main-28.07.2022

Three Dimensional Geometry

121232 The shortest distance between the lines $\frac{x + 7}{- 6} = \frac{y - 6}{7} = z$ and $\frac{7 - x}{2} = y - 2 = z - 6$ is

1

2 \sqrt{29}

2 1

3

\sqrt{\frac{37}{29}}

4

\sqrt{\frac{29}{2}}

Explanation:

A Given,
$l_{1} : \frac{x + 7}{- 6} = \frac{y - 6}{7} = \frac{z - 0}{1}$
And, $l_{2} : \frac{x - 7}{- 2} = \frac{y - 2}{1} = \frac{z - 6}{1}$
$(x_{1}, y_{1}, z_{1}) = (- 7, 6, 0)$
$(a_{1}, b_{1}, c_{1}) = (- 6, 7, 1)$
And, $(x_{2}, y_{2}, z_{2}) = (7, 2, 6)$
$(a_{2}, b_{2}, c_{2}) = (- 2, 1, 1)$
Shortest distance between two lines
$\begin{aligned} d = \frac{| \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} |}{\sqrt{{(a_{1} b_{2} - a_{2} b_{1})}^{2} + {(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2}}} \\ d = \frac{| \begin{array}{ccc} 14 & - 4 & 6 \\ - 6 & 7 & 1 \\ - 2 & 1 & 1 \end{array} |}{\sqrt{(- 6 + 14)^{2} + (7 - 1)^{2} + (- 2 + 6)^{2}}} \end{aligned}$
$\Rightarrow d = | \frac{14 (7 - 1) + 4 (- 6 + 2) + 6 (- 6 + 14)}{\sqrt{64 + 36 + 16}} |$
$\Rightarrow d = \frac{116}{\sqrt{116}} = \sqrt{116}$ Hence, $d = 2 \sqrt{29}$

JEE Main-25.07.2022

Three Dimensional Geometry

121236 If the vectors $α = \hat{i} + a \hat{j} + a^{2} \hat{k}, \boldsymbol β = \hat{i} + b \hat{j} + b^{2} \hat{k}$ and $γ = \hat{i} + c \hat{j} + c^{2} \hat{k}$ are three non- coplanar vectors and $| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{2} \end{array} | = 0$ , then the value of abc is

1 1

2 0

3 -1

4 2

Explanation:

C Given,
$α = \hat{i} + a \hat{j} + a^{2} \hat{k}$
$β = \hat{i} + b \hat{j} + b^{2} \hat{k}$
$γ = \hat{i} + c \hat{j} + c^{2} \hat{k}$
$α, β$ and $γ$ are non-coplanar
So, $abc \neq 0$
Now,
$| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = 0$
We can write as -
$\begin{aligned} | \begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array} | + | \begin{array}{lll} a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3} \end{array} | = 0 \\ \Rightarrow | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | + a b c | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0 \\ \Rightarrow | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | . (1 + abc) = 0 \end{aligned}$
Now, as, $| \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | \neq 0$
Hence,
$\Rightarrow 1 + abc = 0$
$abc = - 1$

WB JEE-2020

Three Dimensional Geometry

121238 The value of ' $x$ ' for which the points $(1, 2, 1)$ . $(0, 1, 3), (1, 0, 1)$ and $(2, 0, x)$ are coplanar is

1

x = 0

2

x = 1

3

x = - 1

4

x = 2

Explanation:

C Given,
$A (1, 2, 1), B (0, 1, 3), C (1, 0, 1)$ and $D (2, 0, x)$ are coplanar
$\vec{AB} = \vec{B} - \vec{A}$
$= (0 \hat{i} + \hat{j} + 3 \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$= - \hat{i} - \hat{j} + 2 \hat{k}$
$\vec{AC} = \vec{C} - \vec{A}$
$= (\hat{i} + 0 \hat{j} + \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$b_{1} \times b_{2} = 12 \hat{i} + 8 \hat{j} - 8 \hat{k}$
$\vec{AD} = \vec{D} - \vec{A}$
$= (2 \hat{i} + 0 \hat{j} + x \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$= \hat{i} - 2 \hat{j} + (x - 1) \hat{k}$
Since, A, B, C and D are coplanar, then
$[\begin{array}{lll} \vec{AB} & \vec{AC} & \vec{AD} \end{array}] = 0$
$[\begin{array}{ccc} - 1 & - 1 & 2 \\ 0 & - 2 & 0 \\ 1 & - 2 & x - 1 \end{array}] = 0$
$\Rightarrow - 1 [- 2 (x - 1)] + 1 (0) + 2 (+ 2) = 0$
$\Rightarrow - (- 2 x + 2) + 0 + 4 = 0$
$\Rightarrow 2 x - 2 + 4 = 0$
$2 x + 2 = 0$ Hence, $x = - 1$

J and K CET-2017

Three Dimensional Geometry

121212 Lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- K}$ and $\frac{x - 1}{K} = \frac{y - 4}{2} =$ $\frac{z - 5}{1}$ are coplanar if

1

K = 2

2

K = 0

3

K = 3

4

K = - 1

Explanation:

B Given,
$\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}} is compared$
with $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- K}$
Then, $x_{1} = 2, y_{1} = 3, z_{1} = 4$
Direction ratios, ( $a_{1} = 1, b_{1} = 1, c_{1} = - K)$
$∴$ And, $\frac{x - 1}{K} = \frac{x - 4}{2} = \frac{z - 5}{1}$
$x_{2} = 1, y_{2} = 4, z_{2} = 5$
Direction ratios $(a_{2} = K, b_{2} = 2, c_{2} = 1)$
The two line will be coplanar If,
$\begin{aligned} | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0 \\ \Rightarrow | \begin{array}{ccc} 1 - 2 & 4 - 3 & 5 - 4 \\ 1 & 1 & - K \\ K & 2 & 1 \end{array} | = 0 \end{aligned}$
$\Rightarrow - 1 (1 + 2 K) - 1 (1 + K^{2}) + 1 (2 - K) = 0$
$\Rightarrow - 1 - 2 K - 1 - K^{2} + 2 - K = 0$
$\Rightarrow - 2 K - K^{2} - K = 0$
$\Rightarrow K^{2} + 3 K = 0$
$\Rightarrow K (K + 3) = 0$
$Hence, K = (0, - 3)$ Hence,

BITSAT-2016

Three Dimensional Geometry

121231 Let the line $\frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lies on P ?

1

(0, - 2, - 2)

2

(- 5, 0, - 1)

3

(3, - 1, 0)

4

(0, 4, 5)

Explanation:

D Given,
$l_{1} : \frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$
$∴$ The point $(1, 2, 3)$ and direction ratio $(λ, 1, 2)$
And, $l_{2} : \frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$
$∵$ The point $(- 26, - 18, - 28)$ and direction ratio $(- 2$ , $3, λ)$
$l_{1}$ and $l_{2}$ are coplanar
We know that, the lines are coplanar
$\Rightarrow | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0$
$\Rightarrow | \begin{array}{ccc} - 27 & - 20 & - 31 \\ λ & 1 & 2 \\ - 2 & 3 & λ \end{array} | = 0$
$\Rightarrow λ^{2} + 9 - 6 λ = 0$
$\Rightarrow λ = \pm 3$
Now, vector perpendicular to plane $P$ which contains given lines is -
$| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ - 2 & 3 & 3 \end{array} | = - 3 \hat{i} - 13 \hat{j} + 11 \hat{k}$
Equation of required plane $P$ is,
$- 3 (x - 1) - 13 (y - 2) + 11 (z - 3) = 0$
$\Rightarrow 3 x + 13 y - 11 z + 4 = 0$
$∴ (0, 4, 5)$ does not lie on plane $P$ .

JEE Main-28.07.2022

Three Dimensional Geometry

121232 The shortest distance between the lines $\frac{x + 7}{- 6} = \frac{y - 6}{7} = z$ and $\frac{7 - x}{2} = y - 2 = z - 6$ is

1

2 \sqrt{29}

2 1

3

\sqrt{\frac{37}{29}}

4

\sqrt{\frac{29}{2}}

Explanation:

A Given,
$l_{1} : \frac{x + 7}{- 6} = \frac{y - 6}{7} = \frac{z - 0}{1}$
And, $l_{2} : \frac{x - 7}{- 2} = \frac{y - 2}{1} = \frac{z - 6}{1}$
$(x_{1}, y_{1}, z_{1}) = (- 7, 6, 0)$
$(a_{1}, b_{1}, c_{1}) = (- 6, 7, 1)$
And, $(x_{2}, y_{2}, z_{2}) = (7, 2, 6)$
$(a_{2}, b_{2}, c_{2}) = (- 2, 1, 1)$
Shortest distance between two lines
$\begin{aligned} d = \frac{| \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} |}{\sqrt{{(a_{1} b_{2} - a_{2} b_{1})}^{2} + {(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2}}} \\ d = \frac{| \begin{array}{ccc} 14 & - 4 & 6 \\ - 6 & 7 & 1 \\ - 2 & 1 & 1 \end{array} |}{\sqrt{(- 6 + 14)^{2} + (7 - 1)^{2} + (- 2 + 6)^{2}}} \end{aligned}$
$\Rightarrow d = | \frac{14 (7 - 1) + 4 (- 6 + 2) + 6 (- 6 + 14)}{\sqrt{64 + 36 + 16}} |$
$\Rightarrow d = \frac{116}{\sqrt{116}} = \sqrt{116}$ Hence, $d = 2 \sqrt{29}$

JEE Main-25.07.2022

Three Dimensional Geometry

121236 If the vectors $α = \hat{i} + a \hat{j} + a^{2} \hat{k}, \boldsymbol β = \hat{i} + b \hat{j} + b^{2} \hat{k}$ and $γ = \hat{i} + c \hat{j} + c^{2} \hat{k}$ are three non- coplanar vectors and $| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{2} \end{array} | = 0$ , then the value of abc is

1 1

2 0

3 -1

4 2

Explanation:

C Given,
$α = \hat{i} + a \hat{j} + a^{2} \hat{k}$
$β = \hat{i} + b \hat{j} + b^{2} \hat{k}$
$γ = \hat{i} + c \hat{j} + c^{2} \hat{k}$
$α, β$ and $γ$ are non-coplanar
So, $abc \neq 0$
Now,
$| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = 0$
We can write as -
$\begin{aligned} | \begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array} | + | \begin{array}{lll} a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3} \end{array} | = 0 \\ \Rightarrow | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | + a b c | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0 \\ \Rightarrow | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | . (1 + abc) = 0 \end{aligned}$
Now, as, $| \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | \neq 0$
Hence,
$\Rightarrow 1 + abc = 0$
$abc = - 1$

WB JEE-2020

Three Dimensional Geometry

121238 The value of ' $x$ ' for which the points $(1, 2, 1)$ . $(0, 1, 3), (1, 0, 1)$ and $(2, 0, x)$ are coplanar is

1

x = 0

2

x = 1

3

x = - 1

4

x = 2

Explanation:

C Given,
$A (1, 2, 1), B (0, 1, 3), C (1, 0, 1)$ and $D (2, 0, x)$ are coplanar
$\vec{AB} = \vec{B} - \vec{A}$
$= (0 \hat{i} + \hat{j} + 3 \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$= - \hat{i} - \hat{j} + 2 \hat{k}$
$\vec{AC} = \vec{C} - \vec{A}$
$= (\hat{i} + 0 \hat{j} + \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$b_{1} \times b_{2} = 12 \hat{i} + 8 \hat{j} - 8 \hat{k}$
$\vec{AD} = \vec{D} - \vec{A}$
$= (2 \hat{i} + 0 \hat{j} + x \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$= \hat{i} - 2 \hat{j} + (x - 1) \hat{k}$
Since, A, B, C and D are coplanar, then
$[\begin{array}{lll} \vec{AB} & \vec{AC} & \vec{AD} \end{array}] = 0$
$[\begin{array}{ccc} - 1 & - 1 & 2 \\ 0 & - 2 & 0 \\ 1 & - 2 & x - 1 \end{array}] = 0$
$\Rightarrow - 1 [- 2 (x - 1)] + 1 (0) + 2 (+ 2) = 0$
$\Rightarrow - (- 2 x + 2) + 0 + 4 = 0$
$\Rightarrow 2 x - 2 + 4 = 0$
$2 x + 2 = 0$ Hence, $x = - 1$

J and K CET-2017

Three Dimensional Geometry

121212 Lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- K}$ and $\frac{x - 1}{K} = \frac{y - 4}{2} =$ $\frac{z - 5}{1}$ are coplanar if

1

K = 2

2

K = 0

3

K = 3

4

K = - 1

Explanation:

B Given,
$\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}} is compared$
with $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- K}$
Then, $x_{1} = 2, y_{1} = 3, z_{1} = 4$
Direction ratios, ( $a_{1} = 1, b_{1} = 1, c_{1} = - K)$
$∴$ And, $\frac{x - 1}{K} = \frac{x - 4}{2} = \frac{z - 5}{1}$
$x_{2} = 1, y_{2} = 4, z_{2} = 5$
Direction ratios $(a_{2} = K, b_{2} = 2, c_{2} = 1)$
The two line will be coplanar If,
$\begin{aligned} | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0 \\ \Rightarrow | \begin{array}{ccc} 1 - 2 & 4 - 3 & 5 - 4 \\ 1 & 1 & - K \\ K & 2 & 1 \end{array} | = 0 \end{aligned}$
$\Rightarrow - 1 (1 + 2 K) - 1 (1 + K^{2}) + 1 (2 - K) = 0$
$\Rightarrow - 1 - 2 K - 1 - K^{2} + 2 - K = 0$
$\Rightarrow - 2 K - K^{2} - K = 0$
$\Rightarrow K^{2} + 3 K = 0$
$\Rightarrow K (K + 3) = 0$
$Hence, K = (0, - 3)$ Hence,

BITSAT-2016

Three Dimensional Geometry

121231 Let the line $\frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lies on P ?

1

(0, - 2, - 2)

2

(- 5, 0, - 1)

3

(3, - 1, 0)

4

(0, 4, 5)

Explanation:

D Given,
$l_{1} : \frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$
$∴$ The point $(1, 2, 3)$ and direction ratio $(λ, 1, 2)$
And, $l_{2} : \frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$
$∵$ The point $(- 26, - 18, - 28)$ and direction ratio $(- 2$ , $3, λ)$
$l_{1}$ and $l_{2}$ are coplanar
We know that, the lines are coplanar
$\Rightarrow | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0$
$\Rightarrow | \begin{array}{ccc} - 27 & - 20 & - 31 \\ λ & 1 & 2 \\ - 2 & 3 & λ \end{array} | = 0$
$\Rightarrow λ^{2} + 9 - 6 λ = 0$
$\Rightarrow λ = \pm 3$
Now, vector perpendicular to plane $P$ which contains given lines is -
$| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ - 2 & 3 & 3 \end{array} | = - 3 \hat{i} - 13 \hat{j} + 11 \hat{k}$
Equation of required plane $P$ is,
$- 3 (x - 1) - 13 (y - 2) + 11 (z - 3) = 0$
$\Rightarrow 3 x + 13 y - 11 z + 4 = 0$
$∴ (0, 4, 5)$ does not lie on plane $P$ .

JEE Main-28.07.2022

Three Dimensional Geometry

121232 The shortest distance between the lines $\frac{x + 7}{- 6} = \frac{y - 6}{7} = z$ and $\frac{7 - x}{2} = y - 2 = z - 6$ is

1

2 \sqrt{29}

2 1

3

\sqrt{\frac{37}{29}}

4

\sqrt{\frac{29}{2}}

Explanation:

A Given,
$l_{1} : \frac{x + 7}{- 6} = \frac{y - 6}{7} = \frac{z - 0}{1}$
And, $l_{2} : \frac{x - 7}{- 2} = \frac{y - 2}{1} = \frac{z - 6}{1}$
$(x_{1}, y_{1}, z_{1}) = (- 7, 6, 0)$
$(a_{1}, b_{1}, c_{1}) = (- 6, 7, 1)$
And, $(x_{2}, y_{2}, z_{2}) = (7, 2, 6)$
$(a_{2}, b_{2}, c_{2}) = (- 2, 1, 1)$
Shortest distance between two lines
$\begin{aligned} d = \frac{| \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} |}{\sqrt{{(a_{1} b_{2} - a_{2} b_{1})}^{2} + {(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2}}} \\ d = \frac{| \begin{array}{ccc} 14 & - 4 & 6 \\ - 6 & 7 & 1 \\ - 2 & 1 & 1 \end{array} |}{\sqrt{(- 6 + 14)^{2} + (7 - 1)^{2} + (- 2 + 6)^{2}}} \end{aligned}$
$\Rightarrow d = | \frac{14 (7 - 1) + 4 (- 6 + 2) + 6 (- 6 + 14)}{\sqrt{64 + 36 + 16}} |$
$\Rightarrow d = \frac{116}{\sqrt{116}} = \sqrt{116}$ Hence, $d = 2 \sqrt{29}$

JEE Main-25.07.2022

Three Dimensional Geometry

121236 If the vectors $α = \hat{i} + a \hat{j} + a^{2} \hat{k}, \boldsymbol β = \hat{i} + b \hat{j} + b^{2} \hat{k}$ and $γ = \hat{i} + c \hat{j} + c^{2} \hat{k}$ are three non- coplanar vectors and $| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{2} \end{array} | = 0$ , then the value of abc is

1 1

2 0

3 -1

4 2

Explanation:

C Given,
$α = \hat{i} + a \hat{j} + a^{2} \hat{k}$
$β = \hat{i} + b \hat{j} + b^{2} \hat{k}$
$γ = \hat{i} + c \hat{j} + c^{2} \hat{k}$
$α, β$ and $γ$ are non-coplanar
So, $abc \neq 0$
Now,
$| \begin{array}{lll} a & a^{2} & 1 + a^{3} \\ b & b^{2} & 1 + b^{3} \\ c & c^{2} & 1 + c^{3} \end{array} | = 0$
We can write as -
$\begin{aligned} | \begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array} | + | \begin{array}{lll} a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3} \end{array} | = 0 \\ \Rightarrow | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | + a b c | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = 0 \\ \Rightarrow | \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | . (1 + abc) = 0 \end{aligned}$
Now, as, $| \begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | \neq 0$
Hence,
$\Rightarrow 1 + abc = 0$
$abc = - 1$

WB JEE-2020

Three Dimensional Geometry

121238 The value of ' $x$ ' for which the points $(1, 2, 1)$ . $(0, 1, 3), (1, 0, 1)$ and $(2, 0, x)$ are coplanar is

1

x = 0

2

x = 1

3

x = - 1

4

x = 2

Explanation:

C Given,
$A (1, 2, 1), B (0, 1, 3), C (1, 0, 1)$ and $D (2, 0, x)$ are coplanar
$\vec{AB} = \vec{B} - \vec{A}$
$= (0 \hat{i} + \hat{j} + 3 \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$= - \hat{i} - \hat{j} + 2 \hat{k}$
$\vec{AC} = \vec{C} - \vec{A}$
$= (\hat{i} + 0 \hat{j} + \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$b_{1} \times b_{2} = 12 \hat{i} + 8 \hat{j} - 8 \hat{k}$
$\vec{AD} = \vec{D} - \vec{A}$
$= (2 \hat{i} + 0 \hat{j} + x \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k})$
$= \hat{i} - 2 \hat{j} + (x - 1) \hat{k}$
Since, A, B, C and D are coplanar, then
$[\begin{array}{lll} \vec{AB} & \vec{AC} & \vec{AD} \end{array}] = 0$
$[\begin{array}{ccc} - 1 & - 1 & 2 \\ 0 & - 2 & 0 \\ 1 & - 2 & x - 1 \end{array}] = 0$
$\Rightarrow - 1 [- 2 (x - 1)] + 1 (0) + 2 (+ 2) = 0$
$\Rightarrow - (- 2 x + 2) + 0 + 4 = 0$
$\Rightarrow 2 x - 2 + 4 = 0$
$2 x + 2 = 0$ Hence, $x = - 1$

J and K CET-2017

Three Dimensional Geometry

121212 Lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- K}$ and $\frac{x - 1}{K} = \frac{y - 4}{2} =$ $\frac{z - 5}{1}$ are coplanar if

1

K = 2

2

K = 0

3

K = 3

4

K = - 1

Explanation:

B Given,
$\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}} is compared$
with $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- K}$
Then, $x_{1} = 2, y_{1} = 3, z_{1} = 4$
Direction ratios, ( $a_{1} = 1, b_{1} = 1, c_{1} = - K)$
$∴$ And, $\frac{x - 1}{K} = \frac{x - 4}{2} = \frac{z - 5}{1}$
$x_{2} = 1, y_{2} = 4, z_{2} = 5$
Direction ratios $(a_{2} = K, b_{2} = 2, c_{2} = 1)$
The two line will be coplanar If,
$\begin{aligned} | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0 \\ \Rightarrow | \begin{array}{ccc} 1 - 2 & 4 - 3 & 5 - 4 \\ 1 & 1 & - K \\ K & 2 & 1 \end{array} | = 0 \end{aligned}$
$\Rightarrow - 1 (1 + 2 K) - 1 (1 + K^{2}) + 1 (2 - K) = 0$
$\Rightarrow - 1 - 2 K - 1 - K^{2} + 2 - K = 0$
$\Rightarrow - 2 K - K^{2} - K = 0$
$\Rightarrow K^{2} + 3 K = 0$
$\Rightarrow K (K + 3) = 0$
$Hence, K = (0, - 3)$ Hence,

BITSAT-2016