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Three Dimensional Geometry

121130 Direction ratios of the line which is perpendicular to the lines with direction ratios $- 1, 2, 2$ and $0, 2, 1$ are

1

1, 1, 2

2

2, - 1, 2

3

- 2, 1, 2

4

2, 1, - 2

Explanation:

B
Let direction ratios of the required line be $a, b, c$
$∴ - 1 a + 2 b + 2 c = 0$
$0 a + 2 b + 1 c = 0$
$∴ \frac{a}{| \begin{array}{cc} 2 & 2 \\ 2 & 1 \end{array} |} = \frac{- b}{| \begin{array}{cc} - 1 & 2 \\ 0 & 1 \end{array} |} = \frac{c}{| \begin{array}{cc} - 1 & 2 \\ 0 & 2 \end{array} |}$
$∴ \frac{a}{2 - 4} = \frac{- b}{- 1 - 0} = \frac{c}{- 2 - 0} \Rightarrow \frac{a}{- 2} = \frac{b}{1} = \frac{c}{- 2} \Rightarrow \frac{a}{2} = \frac{b}{- 1} = \frac{c}{2}$ Hence direction ratios of the required line are $(2, - 1, 2)$

MHT CET-2016

Three Dimensional Geometry

121131 Direction cosines of the line $\frac{x + 2}{2} = \frac{2 y - 5}{3}, z = - 1$ are

1

\frac{4}{5}, \frac{3}{5}, 0

2

\frac{3}{5}, \frac{4}{5}, \frac{1}{5}

3

- \frac{3}{5}, \frac{4}{5}, 0

4

\frac{4}{5}, - \frac{2}{5}, \frac{1}{5}

Explanation:

A
The equation of the line is
$\frac{x - (- 2)}{2} = \frac{y - (\frac{5}{2})}{(\frac{3}{2})}, (z = - 1) i.e$
Hence, the line is passing through the point $(- 2, \frac{5}{2}, - 1)$ and direction ratios of the line are $2, \frac{3}{2}, 0$ i.e. $4, 3, 0$
$∴$ Direction cosines of the line are
$\frac{4}{\sqrt{(4)^{2} + (3)^{2} + 0}}, \frac{3}{\sqrt{(4)^{2} + (3)^{2} + 0}}, \frac{0}{\sqrt{(4)^{2} + (3)^{2} + 0}}$ i.e.
$\frac{4}{5}, \frac{3}{5}, 0$

MHT CET-2016

Three Dimensional Geometry

121133 A plane meets the axes in $A, B$ and $C$ such that centroid of the $△ ABC$ is $(1, 2, 3)$ . The equation of the plane is

1

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

2

\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1

3

x + 2 y + 3 z = 1

4 None of these

Explanation:

B We know that,
Equation of plane
$\frac{x}{x_{1}} + \frac{y}{y_{1}} + \frac{z}{z_{1}} = 1$
$∴$
It meets the coordinate-axes on the points $A (x_{1}, 0, 0), B (0, y_{1}, 0)$ and $C (0, 0, z_{1})$
Then the centroid of $△ ABC$ is-
$= (\frac{x_{1} + 0 + 0}{3}, \frac{0 + y_{1} + y}{3}, \frac{0 + 0 + z_{1}}{3})$
$= (\frac{x_{1}}{3}, \frac{y_{1}}{3}, \frac{z_{1}}{3})$
centroid, $(1, 2, 3) (∵$ Given $)$
Then,
$∵ \frac{x_{1}}{3} = 1, \frac{y_{1}}{3} = 2, \frac{z_{1}}{3} = 3$
$\Rightarrow x_{1} = 3, y_{1} = 6, z_{1} = 9$
Therefore, the equation (i) of the plane becomes-
$\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$

MHT CET-2011

Three Dimensional Geometry

121135 The cosine of the angle $A$ of the triangle with vertices $A (1, - 1, 2), B (6, 11, 2), C (1, 2, 6)$ is

1

\frac{63}{65}

2

\frac{36}{65}

3

\frac{16}{65}

4 none of these

Explanation:

D Given -
original image
$c = AB = \sqrt{(6 - 1)^{2} + (11 + 1)^{2} + (2 - 2)^{2}}$
$= \sqrt{25 + 144} = \sqrt{169}$
$\Rightarrow AB = c = 13$
$and b = AC = \sqrt{(1 - 1)^{2} + (2 + 1)^{2} + (6 - 2)^{2}}$
$\Rightarrow b = \sqrt{0 + 9 + 16} = 5$
And $a = BC = \sqrt{(1 - 6)^{2} + (2 - 11)^{2} + (6 - 2)^{2}}$
$= \sqrt{25 + 81 + 16} = \sqrt{122}$
We know that -
$\cos A = (\frac{b^{2} + c^{2} - a^{2}}{2 bc})$
$= \frac{25 + 169 - 122}{2 \times 5 \times 13} = \frac{194 - 122}{130} = \frac{72}{130} = \frac{36}{65}$

COMEDK-2019

Three Dimensional Geometry

121136 If the angle between the vectors $\vec{a}$ and $\vec{b}$ having direction ratios $1, 2, 1$ and $1, 3 k, 1$ is $\frac{π}{4}$ ,thenk $=$

1

\frac{2 \pm 3 \sqrt{2}}{3}

2

\frac{- 2 \pm 3 \sqrt{2}}{3}

3

\frac{4 \pm 3 \sqrt{2}}{3}

4

\frac{- 4 \pm 3 \sqrt{2}}{3}

Explanation:

D Given -
Vectors $\vec{a}$ and $\vec{b}$ having direction ratio 1,2,1 and 1,3k,1 $∴ x_{1}, y_{1}, z_{1} = 1, 2$ ,
$x_{2}, y_{2}, z_{2} = 1, 3,$
Thus angle between this vectors is given by $θ$
$\cos θ = \frac{x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}}{(\sqrt{x_{1}^{2} + y_{1}^{2} + z_{1}^{2}}) (\sqrt{x_{2}^{2} + y_{2}^{2} + z_{2}^{2}})}$
$as, θ = \frac{π}{4} (∵ given)$
$\cos \frac{π}{4} = \frac{(1) (1) + (2) (3 k) + (1)}{(\sqrt{(1)^{2} + (2)^{2} (1)^{2}}) (\sqrt{1^{2} + (3 k)^{2} + (1)^{2}})}$
$∴ \frac{1}{\sqrt{2}} = \frac{1 + 6 k + 1}{\sqrt{2 + 4} \sqrt{2 + 9 k^{2}}}$
Squaring both sides-
$\frac{1}{2} = \frac{(2 + 6 k)^{2}}{(6) (2 + 9 k^{2})}$
$∴ 3 (2 + 9 k^{2}) = 4 + 36 k^{2} + 24 k$
$6 + 27 k^{2} = 4 + 36 k^{2} + 24 k$
$\Rightarrow 9 k^{2} + 24 k - 2 = 0$
$Thus, k = \frac{- 24 \pm \sqrt{(24)^{2} + 4 (9) (2)}}{18}$
$\Rightarrow k = \frac{- 24 \pm 18 \sqrt{2}}{18}$
$∴ k = \frac{- 4 \pm 3 \sqrt{2}}{3}$

COMEDK-2020

Three Dimensional Geometry

121130 Direction ratios of the line which is perpendicular to the lines with direction ratios $- 1, 2, 2$ and $0, 2, 1$ are

1

1, 1, 2

2

2, - 1, 2

3

- 2, 1, 2

4

2, 1, - 2

Explanation:

B
Let direction ratios of the required line be $a, b, c$
$∴ - 1 a + 2 b + 2 c = 0$
$0 a + 2 b + 1 c = 0$
$∴ \frac{a}{| \begin{array}{cc} 2 & 2 \\ 2 & 1 \end{array} |} = \frac{- b}{| \begin{array}{cc} - 1 & 2 \\ 0 & 1 \end{array} |} = \frac{c}{| \begin{array}{cc} - 1 & 2 \\ 0 & 2 \end{array} |}$
$∴ \frac{a}{2 - 4} = \frac{- b}{- 1 - 0} = \frac{c}{- 2 - 0} \Rightarrow \frac{a}{- 2} = \frac{b}{1} = \frac{c}{- 2} \Rightarrow \frac{a}{2} = \frac{b}{- 1} = \frac{c}{2}$ Hence direction ratios of the required line are $(2, - 1, 2)$

MHT CET-2016

Three Dimensional Geometry

121131 Direction cosines of the line $\frac{x + 2}{2} = \frac{2 y - 5}{3}, z = - 1$ are

1

\frac{4}{5}, \frac{3}{5}, 0

2

\frac{3}{5}, \frac{4}{5}, \frac{1}{5}

3

- \frac{3}{5}, \frac{4}{5}, 0

4

\frac{4}{5}, - \frac{2}{5}, \frac{1}{5}

Explanation:

A
The equation of the line is
$\frac{x - (- 2)}{2} = \frac{y - (\frac{5}{2})}{(\frac{3}{2})}, (z = - 1) i.e$
Hence, the line is passing through the point $(- 2, \frac{5}{2}, - 1)$ and direction ratios of the line are $2, \frac{3}{2}, 0$ i.e. $4, 3, 0$
$∴$ Direction cosines of the line are
$\frac{4}{\sqrt{(4)^{2} + (3)^{2} + 0}}, \frac{3}{\sqrt{(4)^{2} + (3)^{2} + 0}}, \frac{0}{\sqrt{(4)^{2} + (3)^{2} + 0}}$ i.e.
$\frac{4}{5}, \frac{3}{5}, 0$

MHT CET-2016

Three Dimensional Geometry

121133 A plane meets the axes in $A, B$ and $C$ such that centroid of the $△ ABC$ is $(1, 2, 3)$ . The equation of the plane is

1

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

2

\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1

3

x + 2 y + 3 z = 1

4 None of these

Explanation:

B We know that,
Equation of plane
$\frac{x}{x_{1}} + \frac{y}{y_{1}} + \frac{z}{z_{1}} = 1$
$∴$
It meets the coordinate-axes on the points $A (x_{1}, 0, 0), B (0, y_{1}, 0)$ and $C (0, 0, z_{1})$
Then the centroid of $△ ABC$ is-
$= (\frac{x_{1} + 0 + 0}{3}, \frac{0 + y_{1} + y}{3}, \frac{0 + 0 + z_{1}}{3})$
$= (\frac{x_{1}}{3}, \frac{y_{1}}{3}, \frac{z_{1}}{3})$
centroid, $(1, 2, 3) (∵$ Given $)$
Then,
$∵ \frac{x_{1}}{3} = 1, \frac{y_{1}}{3} = 2, \frac{z_{1}}{3} = 3$
$\Rightarrow x_{1} = 3, y_{1} = 6, z_{1} = 9$
Therefore, the equation (i) of the plane becomes-
$\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$

MHT CET-2011

Three Dimensional Geometry

121135 The cosine of the angle $A$ of the triangle with vertices $A (1, - 1, 2), B (6, 11, 2), C (1, 2, 6)$ is

1

\frac{63}{65}

2

\frac{36}{65}

3

\frac{16}{65}

4 none of these

Explanation:

D Given -
original image
$c = AB = \sqrt{(6 - 1)^{2} + (11 + 1)^{2} + (2 - 2)^{2}}$
$= \sqrt{25 + 144} = \sqrt{169}$
$\Rightarrow AB = c = 13$
$and b = AC = \sqrt{(1 - 1)^{2} + (2 + 1)^{2} + (6 - 2)^{2}}$
$\Rightarrow b = \sqrt{0 + 9 + 16} = 5$
And $a = BC = \sqrt{(1 - 6)^{2} + (2 - 11)^{2} + (6 - 2)^{2}}$
$= \sqrt{25 + 81 + 16} = \sqrt{122}$
We know that -
$\cos A = (\frac{b^{2} + c^{2} - a^{2}}{2 bc})$
$= \frac{25 + 169 - 122}{2 \times 5 \times 13} = \frac{194 - 122}{130} = \frac{72}{130} = \frac{36}{65}$

COMEDK-2019

Three Dimensional Geometry

121136 If the angle between the vectors $\vec{a}$ and $\vec{b}$ having direction ratios $1, 2, 1$ and $1, 3 k, 1$ is $\frac{π}{4}$ ,thenk $=$

1

\frac{2 \pm 3 \sqrt{2}}{3}

2

\frac{- 2 \pm 3 \sqrt{2}}{3}

3

\frac{4 \pm 3 \sqrt{2}}{3}

4

\frac{- 4 \pm 3 \sqrt{2}}{3}

Explanation:

D Given -
Vectors $\vec{a}$ and $\vec{b}$ having direction ratio 1,2,1 and 1,3k,1 $∴ x_{1}, y_{1}, z_{1} = 1, 2$ ,
$x_{2}, y_{2}, z_{2} = 1, 3,$
Thus angle between this vectors is given by $θ$
$\cos θ = \frac{x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}}{(\sqrt{x_{1}^{2} + y_{1}^{2} + z_{1}^{2}}) (\sqrt{x_{2}^{2} + y_{2}^{2} + z_{2}^{2}})}$
$as, θ = \frac{π}{4} (∵ given)$
$\cos \frac{π}{4} = \frac{(1) (1) + (2) (3 k) + (1)}{(\sqrt{(1)^{2} + (2)^{2} (1)^{2}}) (\sqrt{1^{2} + (3 k)^{2} + (1)^{2}})}$
$∴ \frac{1}{\sqrt{2}} = \frac{1 + 6 k + 1}{\sqrt{2 + 4} \sqrt{2 + 9 k^{2}}}$
Squaring both sides-
$\frac{1}{2} = \frac{(2 + 6 k)^{2}}{(6) (2 + 9 k^{2})}$
$∴ 3 (2 + 9 k^{2}) = 4 + 36 k^{2} + 24 k$
$6 + 27 k^{2} = 4 + 36 k^{2} + 24 k$
$\Rightarrow 9 k^{2} + 24 k - 2 = 0$
$Thus, k = \frac{- 24 \pm \sqrt{(24)^{2} + 4 (9) (2)}}{18}$
$\Rightarrow k = \frac{- 24 \pm 18 \sqrt{2}}{18}$
$∴ k = \frac{- 4 \pm 3 \sqrt{2}}{3}$

COMEDK-2020

Three Dimensional Geometry

121130 Direction ratios of the line which is perpendicular to the lines with direction ratios $- 1, 2, 2$ and $0, 2, 1$ are

1

1, 1, 2

2

2, - 1, 2

3

- 2, 1, 2

4

2, 1, - 2

Explanation:

B
Let direction ratios of the required line be $a, b, c$
$∴ - 1 a + 2 b + 2 c = 0$
$0 a + 2 b + 1 c = 0$
$∴ \frac{a}{| \begin{array}{cc} 2 & 2 \\ 2 & 1 \end{array} |} = \frac{- b}{| \begin{array}{cc} - 1 & 2 \\ 0 & 1 \end{array} |} = \frac{c}{| \begin{array}{cc} - 1 & 2 \\ 0 & 2 \end{array} |}$
$∴ \frac{a}{2 - 4} = \frac{- b}{- 1 - 0} = \frac{c}{- 2 - 0} \Rightarrow \frac{a}{- 2} = \frac{b}{1} = \frac{c}{- 2} \Rightarrow \frac{a}{2} = \frac{b}{- 1} = \frac{c}{2}$ Hence direction ratios of the required line are $(2, - 1, 2)$

MHT CET-2016

Three Dimensional Geometry

121131 Direction cosines of the line $\frac{x + 2}{2} = \frac{2 y - 5}{3}, z = - 1$ are

1

\frac{4}{5}, \frac{3}{5}, 0

2

\frac{3}{5}, \frac{4}{5}, \frac{1}{5}

3

- \frac{3}{5}, \frac{4}{5}, 0

4

\frac{4}{5}, - \frac{2}{5}, \frac{1}{5}

Explanation:

A
The equation of the line is
$\frac{x - (- 2)}{2} = \frac{y - (\frac{5}{2})}{(\frac{3}{2})}, (z = - 1) i.e$
Hence, the line is passing through the point $(- 2, \frac{5}{2}, - 1)$ and direction ratios of the line are $2, \frac{3}{2}, 0$ i.e. $4, 3, 0$
$∴$ Direction cosines of the line are
$\frac{4}{\sqrt{(4)^{2} + (3)^{2} + 0}}, \frac{3}{\sqrt{(4)^{2} + (3)^{2} + 0}}, \frac{0}{\sqrt{(4)^{2} + (3)^{2} + 0}}$ i.e.
$\frac{4}{5}, \frac{3}{5}, 0$

MHT CET-2016

Three Dimensional Geometry

121133 A plane meets the axes in $A, B$ and $C$ such that centroid of the $△ ABC$ is $(1, 2, 3)$ . The equation of the plane is

1

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

2

\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1

3

x + 2 y + 3 z = 1

4 None of these

Explanation:

B We know that,
Equation of plane
$\frac{x}{x_{1}} + \frac{y}{y_{1}} + \frac{z}{z_{1}} = 1$
$∴$
It meets the coordinate-axes on the points $A (x_{1}, 0, 0), B (0, y_{1}, 0)$ and $C (0, 0, z_{1})$
Then the centroid of $△ ABC$ is-
$= (\frac{x_{1} + 0 + 0}{3}, \frac{0 + y_{1} + y}{3}, \frac{0 + 0 + z_{1}}{3})$
$= (\frac{x_{1}}{3}, \frac{y_{1}}{3}, \frac{z_{1}}{3})$
centroid, $(1, 2, 3) (∵$ Given $)$
Then,
$∵ \frac{x_{1}}{3} = 1, \frac{y_{1}}{3} = 2, \frac{z_{1}}{3} = 3$
$\Rightarrow x_{1} = 3, y_{1} = 6, z_{1} = 9$
Therefore, the equation (i) of the plane becomes-
$\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$

MHT CET-2011

Three Dimensional Geometry

121135 The cosine of the angle $A$ of the triangle with vertices $A (1, - 1, 2), B (6, 11, 2), C (1, 2, 6)$ is

1

\frac{63}{65}

2

\frac{36}{65}

3

\frac{16}{65}

4 none of these

Explanation:

D Given -
original image
$c = AB = \sqrt{(6 - 1)^{2} + (11 + 1)^{2} + (2 - 2)^{2}}$
$= \sqrt{25 + 144} = \sqrt{169}$
$\Rightarrow AB = c = 13$
$and b = AC = \sqrt{(1 - 1)^{2} + (2 + 1)^{2} + (6 - 2)^{2}}$
$\Rightarrow b = \sqrt{0 + 9 + 16} = 5$
And $a = BC = \sqrt{(1 - 6)^{2} + (2 - 11)^{2} + (6 - 2)^{2}}$
$= \sqrt{25 + 81 + 16} = \sqrt{122}$
We know that -
$\cos A = (\frac{b^{2} + c^{2} - a^{2}}{2 bc})$
$= \frac{25 + 169 - 122}{2 \times 5 \times 13} = \frac{194 - 122}{130} = \frac{72}{130} = \frac{36}{65}$

COMEDK-2019

Three Dimensional Geometry

121136 If the angle between the vectors $\vec{a}$ and $\vec{b}$ having direction ratios $1, 2, 1$ and $1, 3 k, 1$ is $\frac{π}{4}$ ,thenk $=$

1

\frac{2 \pm 3 \sqrt{2}}{3}

2

\frac{- 2 \pm 3 \sqrt{2}}{3}

3

\frac{4 \pm 3 \sqrt{2}}{3}

4

\frac{- 4 \pm 3 \sqrt{2}}{3}

Explanation:

D Given -
Vectors $\vec{a}$ and $\vec{b}$ having direction ratio 1,2,1 and 1,3k,1 $∴ x_{1}, y_{1}, z_{1} = 1, 2$ ,
$x_{2}, y_{2}, z_{2} = 1, 3,$
Thus angle between this vectors is given by $θ$
$\cos θ = \frac{x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}}{(\sqrt{x_{1}^{2} + y_{1}^{2} + z_{1}^{2}}) (\sqrt{x_{2}^{2} + y_{2}^{2} + z_{2}^{2}})}$
$as, θ = \frac{π}{4} (∵ given)$
$\cos \frac{π}{4} = \frac{(1) (1) + (2) (3 k) + (1)}{(\sqrt{(1)^{2} + (2)^{2} (1)^{2}}) (\sqrt{1^{2} + (3 k)^{2} + (1)^{2}})}$
$∴ \frac{1}{\sqrt{2}} = \frac{1 + 6 k + 1}{\sqrt{2 + 4} \sqrt{2 + 9 k^{2}}}$
Squaring both sides-
$\frac{1}{2} = \frac{(2 + 6 k)^{2}}{(6) (2 + 9 k^{2})}$
$∴ 3 (2 + 9 k^{2}) = 4 + 36 k^{2} + 24 k$
$6 + 27 k^{2} = 4 + 36 k^{2} + 24 k$
$\Rightarrow 9 k^{2} + 24 k - 2 = 0$
$Thus, k = \frac{- 24 \pm \sqrt{(24)^{2} + 4 (9) (2)}}{18}$
$\Rightarrow k = \frac{- 24 \pm 18 \sqrt{2}}{18}$
$∴ k = \frac{- 4 \pm 3 \sqrt{2}}{3}$

COMEDK-2020

Three Dimensional Geometry

121130 Direction ratios of the line which is perpendicular to the lines with direction ratios $- 1, 2, 2$ and $0, 2, 1$ are

1

1, 1, 2

2

2, - 1, 2

3

- 2, 1, 2

4

2, 1, - 2

Explanation:

B
Let direction ratios of the required line be $a, b, c$
$∴ - 1 a + 2 b + 2 c = 0$
$0 a + 2 b + 1 c = 0$
$∴ \frac{a}{| \begin{array}{cc} 2 & 2 \\ 2 & 1 \end{array} |} = \frac{- b}{| \begin{array}{cc} - 1 & 2 \\ 0 & 1 \end{array} |} = \frac{c}{| \begin{array}{cc} - 1 & 2 \\ 0 & 2 \end{array} |}$
$∴ \frac{a}{2 - 4} = \frac{- b}{- 1 - 0} = \frac{c}{- 2 - 0} \Rightarrow \frac{a}{- 2} = \frac{b}{1} = \frac{c}{- 2} \Rightarrow \frac{a}{2} = \frac{b}{- 1} = \frac{c}{2}$ Hence direction ratios of the required line are $(2, - 1, 2)$

MHT CET-2016

Three Dimensional Geometry

121131 Direction cosines of the line $\frac{x + 2}{2} = \frac{2 y - 5}{3}, z = - 1$ are

1

\frac{4}{5}, \frac{3}{5}, 0

2

\frac{3}{5}, \frac{4}{5}, \frac{1}{5}

3

- \frac{3}{5}, \frac{4}{5}, 0

4

\frac{4}{5}, - \frac{2}{5}, \frac{1}{5}

Explanation:

A
The equation of the line is
$\frac{x - (- 2)}{2} = \frac{y - (\frac{5}{2})}{(\frac{3}{2})}, (z = - 1) i.e$
Hence, the line is passing through the point $(- 2, \frac{5}{2}, - 1)$ and direction ratios of the line are $2, \frac{3}{2}, 0$ i.e. $4, 3, 0$
$∴$ Direction cosines of the line are
$\frac{4}{\sqrt{(4)^{2} + (3)^{2} + 0}}, \frac{3}{\sqrt{(4)^{2} + (3)^{2} + 0}}, \frac{0}{\sqrt{(4)^{2} + (3)^{2} + 0}}$ i.e.
$\frac{4}{5}, \frac{3}{5}, 0$

MHT CET-2016

Three Dimensional Geometry

121133 A plane meets the axes in $A, B$ and $C$ such that centroid of the $△ ABC$ is $(1, 2, 3)$ . The equation of the plane is

1

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

2

\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1

3

x + 2 y + 3 z = 1

4 None of these

Explanation:

B We know that,
Equation of plane
$\frac{x}{x_{1}} + \frac{y}{y_{1}} + \frac{z}{z_{1}} = 1$
$∴$
It meets the coordinate-axes on the points $A (x_{1}, 0, 0), B (0, y_{1}, 0)$ and $C (0, 0, z_{1})$
Then the centroid of $△ ABC$ is-
$= (\frac{x_{1} + 0 + 0}{3}, \frac{0 + y_{1} + y}{3}, \frac{0 + 0 + z_{1}}{3})$
$= (\frac{x_{1}}{3}, \frac{y_{1}}{3}, \frac{z_{1}}{3})$
centroid, $(1, 2, 3) (∵$ Given $)$
Then,
$∵ \frac{x_{1}}{3} = 1, \frac{y_{1}}{3} = 2, \frac{z_{1}}{3} = 3$
$\Rightarrow x_{1} = 3, y_{1} = 6, z_{1} = 9$
Therefore, the equation (i) of the plane becomes-
$\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$

MHT CET-2011

Three Dimensional Geometry

121135 The cosine of the angle $A$ of the triangle with vertices $A (1, - 1, 2), B (6, 11, 2), C (1, 2, 6)$ is

1

\frac{63}{65}

2

\frac{36}{65}

3

\frac{16}{65}

4 none of these

Explanation:

D Given -
original image
$c = AB = \sqrt{(6 - 1)^{2} + (11 + 1)^{2} + (2 - 2)^{2}}$
$= \sqrt{25 + 144} = \sqrt{169}$
$\Rightarrow AB = c = 13$
$and b = AC = \sqrt{(1 - 1)^{2} + (2 + 1)^{2} + (6 - 2)^{2}}$
$\Rightarrow b = \sqrt{0 + 9 + 16} = 5$
And $a = BC = \sqrt{(1 - 6)^{2} + (2 - 11)^{2} + (6 - 2)^{2}}$
$= \sqrt{25 + 81 + 16} = \sqrt{122}$
We know that -
$\cos A = (\frac{b^{2} + c^{2} - a^{2}}{2 bc})$
$= \frac{25 + 169 - 122}{2 \times 5 \times 13} = \frac{194 - 122}{130} = \frac{72}{130} = \frac{36}{65}$

COMEDK-2019

Three Dimensional Geometry

121136 If the angle between the vectors $\vec{a}$ and $\vec{b}$ having direction ratios $1, 2, 1$ and $1, 3 k, 1$ is $\frac{π}{4}$ ,thenk $=$

1

\frac{2 \pm 3 \sqrt{2}}{3}

2

\frac{- 2 \pm 3 \sqrt{2}}{3}

3

\frac{4 \pm 3 \sqrt{2}}{3}

4

\frac{- 4 \pm 3 \sqrt{2}}{3}

Explanation:

D Given -
Vectors $\vec{a}$ and $\vec{b}$ having direction ratio 1,2,1 and 1,3k,1 $∴ x_{1}, y_{1}, z_{1} = 1, 2$ ,
$x_{2}, y_{2}, z_{2} = 1, 3,$
Thus angle between this vectors is given by $θ$
$\cos θ = \frac{x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}}{(\sqrt{x_{1}^{2} + y_{1}^{2} + z_{1}^{2}}) (\sqrt{x_{2}^{2} + y_{2}^{2} + z_{2}^{2}})}$
$as, θ = \frac{π}{4} (∵ given)$
$\cos \frac{π}{4} = \frac{(1) (1) + (2) (3 k) + (1)}{(\sqrt{(1)^{2} + (2)^{2} (1)^{2}}) (\sqrt{1^{2} + (3 k)^{2} + (1)^{2}})}$
$∴ \frac{1}{\sqrt{2}} = \frac{1 + 6 k + 1}{\sqrt{2 + 4} \sqrt{2 + 9 k^{2}}}$
Squaring both sides-
$\frac{1}{2} = \frac{(2 + 6 k)^{2}}{(6) (2 + 9 k^{2})}$
$∴ 3 (2 + 9 k^{2}) = 4 + 36 k^{2} + 24 k$
$6 + 27 k^{2} = 4 + 36 k^{2} + 24 k$
$\Rightarrow 9 k^{2} + 24 k - 2 = 0$
$Thus, k = \frac{- 24 \pm \sqrt{(24)^{2} + 4 (9) (2)}}{18}$
$\Rightarrow k = \frac{- 24 \pm 18 \sqrt{2}}{18}$
$∴ k = \frac{- 4 \pm 3 \sqrt{2}}{3}$

COMEDK-2020

Three Dimensional Geometry

121130 Direction ratios of the line which is perpendicular to the lines with direction ratios $- 1, 2, 2$ and $0, 2, 1$ are

1

1, 1, 2

2

2, - 1, 2

3

- 2, 1, 2

4

2, 1, - 2

Explanation:

B
Let direction ratios of the required line be $a, b, c$
$∴ - 1 a + 2 b + 2 c = 0$
$0 a + 2 b + 1 c = 0$
$∴ \frac{a}{| \begin{array}{cc} 2 & 2 \\ 2 & 1 \end{array} |} = \frac{- b}{| \begin{array}{cc} - 1 & 2 \\ 0 & 1 \end{array} |} = \frac{c}{| \begin{array}{cc} - 1 & 2 \\ 0 & 2 \end{array} |}$
$∴ \frac{a}{2 - 4} = \frac{- b}{- 1 - 0} = \frac{c}{- 2 - 0} \Rightarrow \frac{a}{- 2} = \frac{b}{1} = \frac{c}{- 2} \Rightarrow \frac{a}{2} = \frac{b}{- 1} = \frac{c}{2}$ Hence direction ratios of the required line are $(2, - 1, 2)$

MHT CET-2016

Three Dimensional Geometry

121131 Direction cosines of the line $\frac{x + 2}{2} = \frac{2 y - 5}{3}, z = - 1$ are

1

\frac{4}{5}, \frac{3}{5}, 0

2

\frac{3}{5}, \frac{4}{5}, \frac{1}{5}

3

- \frac{3}{5}, \frac{4}{5}, 0

4

\frac{4}{5}, - \frac{2}{5}, \frac{1}{5}

Explanation:

A
The equation of the line is
$\frac{x - (- 2)}{2} = \frac{y - (\frac{5}{2})}{(\frac{3}{2})}, (z = - 1) i.e$
Hence, the line is passing through the point $(- 2, \frac{5}{2}, - 1)$ and direction ratios of the line are $2, \frac{3}{2}, 0$ i.e. $4, 3, 0$
$∴$ Direction cosines of the line are
$\frac{4}{\sqrt{(4)^{2} + (3)^{2} + 0}}, \frac{3}{\sqrt{(4)^{2} + (3)^{2} + 0}}, \frac{0}{\sqrt{(4)^{2} + (3)^{2} + 0}}$ i.e.
$\frac{4}{5}, \frac{3}{5}, 0$

MHT CET-2016

Three Dimensional Geometry

121133 A plane meets the axes in $A, B$ and $C$ such that centroid of the $△ ABC$ is $(1, 2, 3)$ . The equation of the plane is

1

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

2

\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1

3

x + 2 y + 3 z = 1

4 None of these

Explanation:

B We know that,
Equation of plane
$\frac{x}{x_{1}} + \frac{y}{y_{1}} + \frac{z}{z_{1}} = 1$
$∴$
It meets the coordinate-axes on the points $A (x_{1}, 0, 0), B (0, y_{1}, 0)$ and $C (0, 0, z_{1})$
Then the centroid of $△ ABC$ is-
$= (\frac{x_{1} + 0 + 0}{3}, \frac{0 + y_{1} + y}{3}, \frac{0 + 0 + z_{1}}{3})$
$= (\frac{x_{1}}{3}, \frac{y_{1}}{3}, \frac{z_{1}}{3})$
centroid, $(1, 2, 3) (∵$ Given $)$
Then,
$∵ \frac{x_{1}}{3} = 1, \frac{y_{1}}{3} = 2, \frac{z_{1}}{3} = 3$
$\Rightarrow x_{1} = 3, y_{1} = 6, z_{1} = 9$
Therefore, the equation (i) of the plane becomes-
$\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$

MHT CET-2011

Three Dimensional Geometry

121135 The cosine of the angle $A$ of the triangle with vertices $A (1, - 1, 2), B (6, 11, 2), C (1, 2, 6)$ is

1

\frac{63}{65}

2

\frac{36}{65}

3

\frac{16}{65}

4 none of these

Explanation:

D Given -
original image
$c = AB = \sqrt{(6 - 1)^{2} + (11 + 1)^{2} + (2 - 2)^{2}}$
$= \sqrt{25 + 144} = \sqrt{169}$
$\Rightarrow AB = c = 13$
$and b = AC = \sqrt{(1 - 1)^{2} + (2 + 1)^{2} + (6 - 2)^{2}}$
$\Rightarrow b = \sqrt{0 + 9 + 16} = 5$
And $a = BC = \sqrt{(1 - 6)^{2} + (2 - 11)^{2} + (6 - 2)^{2}}$
$= \sqrt{25 + 81 + 16} = \sqrt{122}$
We know that -
$\cos A = (\frac{b^{2} + c^{2} - a^{2}}{2 bc})$
$= \frac{25 + 169 - 122}{2 \times 5 \times 13} = \frac{194 - 122}{130} = \frac{72}{130} = \frac{36}{65}$

COMEDK-2019

Three Dimensional Geometry

121136 If the angle between the vectors $\vec{a}$ and $\vec{b}$ having direction ratios $1, 2, 1$ and $1, 3 k, 1$ is $\frac{π}{4}$ ,thenk $=$

1

\frac{2 \pm 3 \sqrt{2}}{3}

2

\frac{- 2 \pm 3 \sqrt{2}}{3}

3

\frac{4 \pm 3 \sqrt{2}}{3}

4

\frac{- 4 \pm 3 \sqrt{2}}{3}

Explanation:

D Given -
Vectors $\vec{a}$ and $\vec{b}$ having direction ratio 1,2,1 and 1,3k,1 $∴ x_{1}, y_{1}, z_{1} = 1, 2$ ,
$x_{2}, y_{2}, z_{2} = 1, 3,$
Thus angle between this vectors is given by $θ$
$\cos θ = \frac{x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}}{(\sqrt{x_{1}^{2} + y_{1}^{2} + z_{1}^{2}}) (\sqrt{x_{2}^{2} + y_{2}^{2} + z_{2}^{2}})}$
$as, θ = \frac{π}{4} (∵ given)$
$\cos \frac{π}{4} = \frac{(1) (1) + (2) (3 k) + (1)}{(\sqrt{(1)^{2} + (2)^{2} (1)^{2}}) (\sqrt{1^{2} + (3 k)^{2} + (1)^{2}})}$
$∴ \frac{1}{\sqrt{2}} = \frac{1 + 6 k + 1}{\sqrt{2 + 4} \sqrt{2 + 9 k^{2}}}$
Squaring both sides-
$\frac{1}{2} = \frac{(2 + 6 k)^{2}}{(6) (2 + 9 k^{2})}$
$∴ 3 (2 + 9 k^{2}) = 4 + 36 k^{2} + 24 k$
$6 + 27 k^{2} = 4 + 36 k^{2} + 24 k$
$\Rightarrow 9 k^{2} + 24 k - 2 = 0$
$Thus, k = \frac{- 24 \pm \sqrt{(24)^{2} + 4 (9) (2)}}{18}$
$\Rightarrow k = \frac{- 24 \pm 18 \sqrt{2}}{18}$
$∴ k = \frac{- 4 \pm 3 \sqrt{2}}{3}$

COMEDK-2020