QBManager

Three Dimensional Geometry

121335 Find the angle between the planes $x + 2 y + 2 z -$ $5 = 0$ and $3 x + 3 y + 2 z - 8 = 0$

1

\cos^{- 1} (\frac{3}{\sqrt{22}})

2

\cos^{- 1} (\frac{13}{3 \sqrt{22}})

3

\cos^{- 1} (\frac{1}{3 \sqrt{22}})

4

\cos^{- 1} (\frac{13}{31})

Explanation:

B Given,
$P_{1} : x + 2 y + 2 z - 5 = 0$
$P_{2} : 3 x + 3 y + 2 z - 8 = 0$
From equation (i) and (ii) we get-
${\vec{n}}_{1} = \hat{i} + 2 \hat{j} + 2 \hat{k}$
${\vec{n}}_{2} = 3 \hat{i} + 3 \hat{j} + 2 \hat{k}$
Angle between planes $P_{1}$ and $P_{2}$ is
$\cos θ = | \frac{({\vec{n}}_{1}) \cdot ({\vec{n}}_{2})}{| {\vec{n}}_{1} | \cdot | {\vec{n}}_{2} |} |$
$\cos θ = | \frac{(\hat{i} + 2 \hat{j} + 2 \hat{k}) \cdot (3 \hat{i} + 3 \hat{j} + 2 \hat{k})}{\sqrt{1 + 4 + 4} \cdot \sqrt{9 + 9 + 4}} |$
$\cos θ = | \frac{3 + 6 + 4}{\sqrt{9} \cdot \sqrt{22}} | = \frac{13}{3 \sqrt{22}}$
$θ = \cos^{- 1} (\frac{13}{3 \sqrt{22}})$

AP EAMCET-22.09.2020

Three Dimensional Geometry

121336 The foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $π_{1}$ is $(1, 3, 5)$ . If $(2, 2$ , $- 1), (3, 4, 2), (3, 4, 2) (3, 3, 0)$ are three points on the plane $π_{2}$ , then the angle between the planes $π_{1}$ and $π_{2}$ is

1

\frac{π}{2}

2

\cos^{- 1} (\frac{1}{3})

3

\frac{π}{6}

4

\cos^{- 1} (\frac{2}{5})

Explanation:

A We know that, foot of perpendicular from a point $(x_{1}, y_{1}, z_{1})$ on the plane $a x + b y + c z + d = 0$ is $(x, y, z)$ where
$\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c} = \frac{- (a_{1} + b y_{1} + c z_{1} + d)}{a^{2} + b^{2} + c^{2}}$
Since, the foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $π_{1}$ is $(1, 3, 5)$
$∴ \frac{1 - 1}{a} = \frac{3 - 1}{b} = \frac{5 - 1}{c} = \frac{- (a + b + c + d)}{a^{2} + b^{2} + c^{2}}$
$\Rightarrow \frac{0}{a} = \frac{2}{b} = \frac{4}{c} = \frac{- (a + b + c + d)}{a^{2} + b^{2} + c^{2}}$
$\Rightarrow a = 0, b = 2 k, c = 4 k$
Equation of plane $π_{1}$ through point $(1, 3, 5)$ and having directions of normal to plane are $a, b, c$ is
$0 (x - 1) + 2 k (y - 3) + 4 k (z - 5) = 0$
$\Rightarrow (y - 3) + 2 (z - 5) = 0$
$\Rightarrow y + 2 z - 13 = 0$
And equation of plane $π_{2}$ is
$\begin{aligned} | \begin{array}{ccc} x - 2 & y - 2 & z + 1 \\ 3 - 2 & 4 - 2 & 2 + 1 \\ 3 - 2 & 3 - 2 & 0 + 1 \end{array} | = 0 \\ | \begin{array}{ccc} x - 2 & y - 2 & z + 1 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{array} | = 0 \end{aligned}$
$\Rightarrow (x - 2) [2 - 3] - (y - 2) [1 - 3] + (z + 1) [1 - 2] = 0$
$\Rightarrow - (x - 2) + 2 (y - 2) - 1 (z + 1) = 0$
$\Rightarrow (x - 2) - 2 (y - 2) + (z + 1) = 0$
$\Rightarrow x - 2 y + z + 3 = 0$
$∴$ Angle between plane $π_{1}$ and $π_{2}$
$\cos θ = | (\frac{0 \times 1 + 1 \times - 2 + 2 \times 1}{\sqrt{0 + 1 + 4} \sqrt{1 + 4 + 1}}) |$
$\cos θ = 0$
$\cos θ = \cos \frac{π}{2}$
$θ = \frac{π}{2}$

TS EAMCET-10.09.2020

Three Dimensional Geometry

121337 A tetrahedron has vertices $O (0, 0, 0), (A (1, 2$ , 1) $B (2, 1, 3), C (- 1, 1, 2)$ . If $θ$ is the angle between the faces $O A B$ and $A B C$ , then $\cos θ =$

1

\frac{1}{\sqrt{2}}

2

\frac{19}{35}

3

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

4

\frac{17}{31}

Explanation:

B Equation of plane OAB is given by,
$| \begin{array}{ccc} x - 0 & y - 0 & z - 0 \\ 1 - 0 & 2 - 0 & 1 - 0 \\ 2 - 0 & 1 - 0 & 3 - 0 \end{array} | = 0 \Rightarrow | \begin{array}{ccc} x & y & z \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array} | = 0$
$x (6 - 1) - y (3 - 2) + z (1 - 4) = 0$
$5 x - y - 3 z = 0$
Equation of plane $ABC$ is given by,
$\begin{aligned} | \begin{array}{ccc} x - 1 & y - 2 & z - 1 \\ 2 - 1 & 1 - 2 & 3 - 1 \\ - 1 - 1 & 1 - 2 & 2 - 1 \end{array} | = 0 \\ | \begin{array}{ccc} x - 1 & y - 2 & z - 1 \\ 1 & - 1 & 2 \\ - 2 & - 1 & 1 \end{array} | = 0 \end{aligned}$
$(x - 1) (- 1 + 2) - (y - 2) (1 + 4) + (z - 1) (- 1 - 2) = 0$
$x - 1 - 5 y + 10 - 3 z + 3 = 0$
$x - 5 y - 3 z + 12 = 0$
Then angle between planes represented by Eqs. (i) and (ii) is given by -
$\cos θ = \frac{(5) (1) + (- 1) (- 5) + (- 3) (- 3)}{\sqrt{25 + 1 + 9} \sqrt{1 + 25 + 9}} = \frac{19}{35}$

TS EAMCET-10.09.2020

Three Dimensional Geometry

121307 If a line makes angles of measure $\frac{π}{6}$ and $\frac{π}{3}$ with $X$ and $Y$ axes respectively, then the angle made by the line with $Z$ axis is

1

\frac{π}{6}

2

\frac{π}{2}

3

\frac{π}{5}

4

\frac{π}{4}

Explanation:

B Let,
| x-axis, | y-axis | z-axis |
| :---: | :---: | :---: |
| $α$ | $β$ | $γ$ |
| $\cos α = 1$ | $\cos β = m$ | $\cos γ = n$ |
We know that,
$l^{2} + m^{2} + n^{2} = 1$
${[\cos (\frac{π}{6})]}^{2} + {[\cos (\frac{π}{3})]}^{2} + (\cos α)^{2} = 1$
$(\frac{3}{4}) + (\frac{1}{4}) + \cos^{2} α = 1$
$1 + \cos^{2} α = 1$
$\cos^{2} α = 0$
$α = \frac{π}{2}$

MHT CET-2020

Three Dimensional Geometry

121335 Find the angle between the planes $x + 2 y + 2 z -$ $5 = 0$ and $3 x + 3 y + 2 z - 8 = 0$

1

\cos^{- 1} (\frac{3}{\sqrt{22}})

2

\cos^{- 1} (\frac{13}{3 \sqrt{22}})

3

\cos^{- 1} (\frac{1}{3 \sqrt{22}})

4

\cos^{- 1} (\frac{13}{31})

Explanation:

B Given,
$P_{1} : x + 2 y + 2 z - 5 = 0$
$P_{2} : 3 x + 3 y + 2 z - 8 = 0$
From equation (i) and (ii) we get-
${\vec{n}}_{1} = \hat{i} + 2 \hat{j} + 2 \hat{k}$
${\vec{n}}_{2} = 3 \hat{i} + 3 \hat{j} + 2 \hat{k}$
Angle between planes $P_{1}$ and $P_{2}$ is
$\cos θ = | \frac{({\vec{n}}_{1}) \cdot ({\vec{n}}_{2})}{| {\vec{n}}_{1} | \cdot | {\vec{n}}_{2} |} |$
$\cos θ = | \frac{(\hat{i} + 2 \hat{j} + 2 \hat{k}) \cdot (3 \hat{i} + 3 \hat{j} + 2 \hat{k})}{\sqrt{1 + 4 + 4} \cdot \sqrt{9 + 9 + 4}} |$
$\cos θ = | \frac{3 + 6 + 4}{\sqrt{9} \cdot \sqrt{22}} | = \frac{13}{3 \sqrt{22}}$
$θ = \cos^{- 1} (\frac{13}{3 \sqrt{22}})$

AP EAMCET-22.09.2020

Three Dimensional Geometry

121336 The foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $π_{1}$ is $(1, 3, 5)$ . If $(2, 2$ , $- 1), (3, 4, 2), (3, 4, 2) (3, 3, 0)$ are three points on the plane $π_{2}$ , then the angle between the planes $π_{1}$ and $π_{2}$ is

1

\frac{π}{2}

2

\cos^{- 1} (\frac{1}{3})

3

\frac{π}{6}

4

\cos^{- 1} (\frac{2}{5})

Explanation:

A We know that, foot of perpendicular from a point $(x_{1}, y_{1}, z_{1})$ on the plane $a x + b y + c z + d = 0$ is $(x, y, z)$ where
$\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c} = \frac{- (a_{1} + b y_{1} + c z_{1} + d)}{a^{2} + b^{2} + c^{2}}$
Since, the foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $π_{1}$ is $(1, 3, 5)$
$∴ \frac{1 - 1}{a} = \frac{3 - 1}{b} = \frac{5 - 1}{c} = \frac{- (a + b + c + d)}{a^{2} + b^{2} + c^{2}}$
$\Rightarrow \frac{0}{a} = \frac{2}{b} = \frac{4}{c} = \frac{- (a + b + c + d)}{a^{2} + b^{2} + c^{2}}$
$\Rightarrow a = 0, b = 2 k, c = 4 k$
Equation of plane $π_{1}$ through point $(1, 3, 5)$ and having directions of normal to plane are $a, b, c$ is
$0 (x - 1) + 2 k (y - 3) + 4 k (z - 5) = 0$
$\Rightarrow (y - 3) + 2 (z - 5) = 0$
$\Rightarrow y + 2 z - 13 = 0$
And equation of plane $π_{2}$ is
$\begin{aligned} | \begin{array}{ccc} x - 2 & y - 2 & z + 1 \\ 3 - 2 & 4 - 2 & 2 + 1 \\ 3 - 2 & 3 - 2 & 0 + 1 \end{array} | = 0 \\ | \begin{array}{ccc} x - 2 & y - 2 & z + 1 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{array} | = 0 \end{aligned}$
$\Rightarrow (x - 2) [2 - 3] - (y - 2) [1 - 3] + (z + 1) [1 - 2] = 0$
$\Rightarrow - (x - 2) + 2 (y - 2) - 1 (z + 1) = 0$
$\Rightarrow (x - 2) - 2 (y - 2) + (z + 1) = 0$
$\Rightarrow x - 2 y + z + 3 = 0$
$∴$ Angle between plane $π_{1}$ and $π_{2}$
$\cos θ = | (\frac{0 \times 1 + 1 \times - 2 + 2 \times 1}{\sqrt{0 + 1 + 4} \sqrt{1 + 4 + 1}}) |$
$\cos θ = 0$
$\cos θ = \cos \frac{π}{2}$
$θ = \frac{π}{2}$

TS EAMCET-10.09.2020

Three Dimensional Geometry

121337 A tetrahedron has vertices $O (0, 0, 0), (A (1, 2$ , 1) $B (2, 1, 3), C (- 1, 1, 2)$ . If $θ$ is the angle between the faces $O A B$ and $A B C$ , then $\cos θ =$

1

\frac{1}{\sqrt{2}}

2

\frac{19}{35}

3

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

4

\frac{17}{31}

Explanation:

B Equation of plane OAB is given by,
$| \begin{array}{ccc} x - 0 & y - 0 & z - 0 \\ 1 - 0 & 2 - 0 & 1 - 0 \\ 2 - 0 & 1 - 0 & 3 - 0 \end{array} | = 0 \Rightarrow | \begin{array}{ccc} x & y & z \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array} | = 0$
$x (6 - 1) - y (3 - 2) + z (1 - 4) = 0$
$5 x - y - 3 z = 0$
Equation of plane $ABC$ is given by,
$\begin{aligned} | \begin{array}{ccc} x - 1 & y - 2 & z - 1 \\ 2 - 1 & 1 - 2 & 3 - 1 \\ - 1 - 1 & 1 - 2 & 2 - 1 \end{array} | = 0 \\ | \begin{array}{ccc} x - 1 & y - 2 & z - 1 \\ 1 & - 1 & 2 \\ - 2 & - 1 & 1 \end{array} | = 0 \end{aligned}$
$(x - 1) (- 1 + 2) - (y - 2) (1 + 4) + (z - 1) (- 1 - 2) = 0$
$x - 1 - 5 y + 10 - 3 z + 3 = 0$
$x - 5 y - 3 z + 12 = 0$
Then angle between planes represented by Eqs. (i) and (ii) is given by -
$\cos θ = \frac{(5) (1) + (- 1) (- 5) + (- 3) (- 3)}{\sqrt{25 + 1 + 9} \sqrt{1 + 25 + 9}} = \frac{19}{35}$

TS EAMCET-10.09.2020

Three Dimensional Geometry

121307 If a line makes angles of measure $\frac{π}{6}$ and $\frac{π}{3}$ with $X$ and $Y$ axes respectively, then the angle made by the line with $Z$ axis is

1

\frac{π}{6}

2

\frac{π}{2}

3

\frac{π}{5}

4

\frac{π}{4}

Explanation:

B Let,
| x-axis, | y-axis | z-axis |
| :---: | :---: | :---: |
| $α$ | $β$ | $γ$ |
| $\cos α = 1$ | $\cos β = m$ | $\cos γ = n$ |
We know that,
$l^{2} + m^{2} + n^{2} = 1$
${[\cos (\frac{π}{6})]}^{2} + {[\cos (\frac{π}{3})]}^{2} + (\cos α)^{2} = 1$
$(\frac{3}{4}) + (\frac{1}{4}) + \cos^{2} α = 1$
$1 + \cos^{2} α = 1$
$\cos^{2} α = 0$
$α = \frac{π}{2}$

MHT CET-2020

Three Dimensional Geometry

121335 Find the angle between the planes $x + 2 y + 2 z -$ $5 = 0$ and $3 x + 3 y + 2 z - 8 = 0$

1

\cos^{- 1} (\frac{3}{\sqrt{22}})

2

\cos^{- 1} (\frac{13}{3 \sqrt{22}})

3

\cos^{- 1} (\frac{1}{3 \sqrt{22}})

4

\cos^{- 1} (\frac{13}{31})

Explanation:

B Given,
$P_{1} : x + 2 y + 2 z - 5 = 0$
$P_{2} : 3 x + 3 y + 2 z - 8 = 0$
From equation (i) and (ii) we get-
${\vec{n}}_{1} = \hat{i} + 2 \hat{j} + 2 \hat{k}$
${\vec{n}}_{2} = 3 \hat{i} + 3 \hat{j} + 2 \hat{k}$
Angle between planes $P_{1}$ and $P_{2}$ is
$\cos θ = | \frac{({\vec{n}}_{1}) \cdot ({\vec{n}}_{2})}{| {\vec{n}}_{1} | \cdot | {\vec{n}}_{2} |} |$
$\cos θ = | \frac{(\hat{i} + 2 \hat{j} + 2 \hat{k}) \cdot (3 \hat{i} + 3 \hat{j} + 2 \hat{k})}{\sqrt{1 + 4 + 4} \cdot \sqrt{9 + 9 + 4}} |$
$\cos θ = | \frac{3 + 6 + 4}{\sqrt{9} \cdot \sqrt{22}} | = \frac{13}{3 \sqrt{22}}$
$θ = \cos^{- 1} (\frac{13}{3 \sqrt{22}})$

AP EAMCET-22.09.2020

Three Dimensional Geometry

121336 The foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $π_{1}$ is $(1, 3, 5)$ . If $(2, 2$ , $- 1), (3, 4, 2), (3, 4, 2) (3, 3, 0)$ are three points on the plane $π_{2}$ , then the angle between the planes $π_{1}$ and $π_{2}$ is

1

\frac{π}{2}

2

\cos^{- 1} (\frac{1}{3})

3

\frac{π}{6}

4

\cos^{- 1} (\frac{2}{5})

Explanation:

A We know that, foot of perpendicular from a point $(x_{1}, y_{1}, z_{1})$ on the plane $a x + b y + c z + d = 0$ is $(x, y, z)$ where
$\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c} = \frac{- (a_{1} + b y_{1} + c z_{1} + d)}{a^{2} + b^{2} + c^{2}}$
Since, the foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $π_{1}$ is $(1, 3, 5)$
$∴ \frac{1 - 1}{a} = \frac{3 - 1}{b} = \frac{5 - 1}{c} = \frac{- (a + b + c + d)}{a^{2} + b^{2} + c^{2}}$
$\Rightarrow \frac{0}{a} = \frac{2}{b} = \frac{4}{c} = \frac{- (a + b + c + d)}{a^{2} + b^{2} + c^{2}}$
$\Rightarrow a = 0, b = 2 k, c = 4 k$
Equation of plane $π_{1}$ through point $(1, 3, 5)$ and having directions of normal to plane are $a, b, c$ is
$0 (x - 1) + 2 k (y - 3) + 4 k (z - 5) = 0$
$\Rightarrow (y - 3) + 2 (z - 5) = 0$
$\Rightarrow y + 2 z - 13 = 0$
And equation of plane $π_{2}$ is
$\begin{aligned} | \begin{array}{ccc} x - 2 & y - 2 & z + 1 \\ 3 - 2 & 4 - 2 & 2 + 1 \\ 3 - 2 & 3 - 2 & 0 + 1 \end{array} | = 0 \\ | \begin{array}{ccc} x - 2 & y - 2 & z + 1 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{array} | = 0 \end{aligned}$
$\Rightarrow (x - 2) [2 - 3] - (y - 2) [1 - 3] + (z + 1) [1 - 2] = 0$
$\Rightarrow - (x - 2) + 2 (y - 2) - 1 (z + 1) = 0$
$\Rightarrow (x - 2) - 2 (y - 2) + (z + 1) = 0$
$\Rightarrow x - 2 y + z + 3 = 0$
$∴$ Angle between plane $π_{1}$ and $π_{2}$
$\cos θ = | (\frac{0 \times 1 + 1 \times - 2 + 2 \times 1}{\sqrt{0 + 1 + 4} \sqrt{1 + 4 + 1}}) |$
$\cos θ = 0$
$\cos θ = \cos \frac{π}{2}$
$θ = \frac{π}{2}$

TS EAMCET-10.09.2020

Three Dimensional Geometry

121337 A tetrahedron has vertices $O (0, 0, 0), (A (1, 2$ , 1) $B (2, 1, 3), C (- 1, 1, 2)$ . If $θ$ is the angle between the faces $O A B$ and $A B C$ , then $\cos θ =$

1

\frac{1}{\sqrt{2}}

2

\frac{19}{35}

3

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

4

\frac{17}{31}

Explanation:

B Equation of plane OAB is given by,
$| \begin{array}{ccc} x - 0 & y - 0 & z - 0 \\ 1 - 0 & 2 - 0 & 1 - 0 \\ 2 - 0 & 1 - 0 & 3 - 0 \end{array} | = 0 \Rightarrow | \begin{array}{ccc} x & y & z \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array} | = 0$
$x (6 - 1) - y (3 - 2) + z (1 - 4) = 0$
$5 x - y - 3 z = 0$
Equation of plane $ABC$ is given by,
$\begin{aligned} | \begin{array}{ccc} x - 1 & y - 2 & z - 1 \\ 2 - 1 & 1 - 2 & 3 - 1 \\ - 1 - 1 & 1 - 2 & 2 - 1 \end{array} | = 0 \\ | \begin{array}{ccc} x - 1 & y - 2 & z - 1 \\ 1 & - 1 & 2 \\ - 2 & - 1 & 1 \end{array} | = 0 \end{aligned}$
$(x - 1) (- 1 + 2) - (y - 2) (1 + 4) + (z - 1) (- 1 - 2) = 0$
$x - 1 - 5 y + 10 - 3 z + 3 = 0$
$x - 5 y - 3 z + 12 = 0$
Then angle between planes represented by Eqs. (i) and (ii) is given by -
$\cos θ = \frac{(5) (1) + (- 1) (- 5) + (- 3) (- 3)}{\sqrt{25 + 1 + 9} \sqrt{1 + 25 + 9}} = \frac{19}{35}$

TS EAMCET-10.09.2020

Three Dimensional Geometry

121307 If a line makes angles of measure $\frac{π}{6}$ and $\frac{π}{3}$ with $X$ and $Y$ axes respectively, then the angle made by the line with $Z$ axis is

1

\frac{π}{6}

2

\frac{π}{2}

3

\frac{π}{5}

4

\frac{π}{4}

Explanation:

B Let,
| x-axis, | y-axis | z-axis |
| :---: | :---: | :---: |
| $α$ | $β$ | $γ$ |
| $\cos α = 1$ | $\cos β = m$ | $\cos γ = n$ |
We know that,
$l^{2} + m^{2} + n^{2} = 1$
${[\cos (\frac{π}{6})]}^{2} + {[\cos (\frac{π}{3})]}^{2} + (\cos α)^{2} = 1$
$(\frac{3}{4}) + (\frac{1}{4}) + \cos^{2} α = 1$
$1 + \cos^{2} α = 1$
$\cos^{2} α = 0$
$α = \frac{π}{2}$

MHT CET-2020

Three Dimensional Geometry

121335 Find the angle between the planes $x + 2 y + 2 z -$ $5 = 0$ and $3 x + 3 y + 2 z - 8 = 0$

1

\cos^{- 1} (\frac{3}{\sqrt{22}})

2

\cos^{- 1} (\frac{13}{3 \sqrt{22}})

3

\cos^{- 1} (\frac{1}{3 \sqrt{22}})

4

\cos^{- 1} (\frac{13}{31})

Explanation:

B Given,
$P_{1} : x + 2 y + 2 z - 5 = 0$
$P_{2} : 3 x + 3 y + 2 z - 8 = 0$
From equation (i) and (ii) we get-
${\vec{n}}_{1} = \hat{i} + 2 \hat{j} + 2 \hat{k}$
${\vec{n}}_{2} = 3 \hat{i} + 3 \hat{j} + 2 \hat{k}$
Angle between planes $P_{1}$ and $P_{2}$ is
$\cos θ = | \frac{({\vec{n}}_{1}) \cdot ({\vec{n}}_{2})}{| {\vec{n}}_{1} | \cdot | {\vec{n}}_{2} |} |$
$\cos θ = | \frac{(\hat{i} + 2 \hat{j} + 2 \hat{k}) \cdot (3 \hat{i} + 3 \hat{j} + 2 \hat{k})}{\sqrt{1 + 4 + 4} \cdot \sqrt{9 + 9 + 4}} |$
$\cos θ = | \frac{3 + 6 + 4}{\sqrt{9} \cdot \sqrt{22}} | = \frac{13}{3 \sqrt{22}}$
$θ = \cos^{- 1} (\frac{13}{3 \sqrt{22}})$

AP EAMCET-22.09.2020

Three Dimensional Geometry

121336 The foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $π_{1}$ is $(1, 3, 5)$ . If $(2, 2$ , $- 1), (3, 4, 2), (3, 4, 2) (3, 3, 0)$ are three points on the plane $π_{2}$ , then the angle between the planes $π_{1}$ and $π_{2}$ is

1

\frac{π}{2}

2

\cos^{- 1} (\frac{1}{3})

3

\frac{π}{6}

4

\cos^{- 1} (\frac{2}{5})

Explanation:

A We know that, foot of perpendicular from a point $(x_{1}, y_{1}, z_{1})$ on the plane $a x + b y + c z + d = 0$ is $(x, y, z)$ where
$\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c} = \frac{- (a_{1} + b y_{1} + c z_{1} + d)}{a^{2} + b^{2} + c^{2}}$
Since, the foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $π_{1}$ is $(1, 3, 5)$
$∴ \frac{1 - 1}{a} = \frac{3 - 1}{b} = \frac{5 - 1}{c} = \frac{- (a + b + c + d)}{a^{2} + b^{2} + c^{2}}$
$\Rightarrow \frac{0}{a} = \frac{2}{b} = \frac{4}{c} = \frac{- (a + b + c + d)}{a^{2} + b^{2} + c^{2}}$
$\Rightarrow a = 0, b = 2 k, c = 4 k$
Equation of plane $π_{1}$ through point $(1, 3, 5)$ and having directions of normal to plane are $a, b, c$ is
$0 (x - 1) + 2 k (y - 3) + 4 k (z - 5) = 0$
$\Rightarrow (y - 3) + 2 (z - 5) = 0$
$\Rightarrow y + 2 z - 13 = 0$
And equation of plane $π_{2}$ is
$\begin{aligned} | \begin{array}{ccc} x - 2 & y - 2 & z + 1 \\ 3 - 2 & 4 - 2 & 2 + 1 \\ 3 - 2 & 3 - 2 & 0 + 1 \end{array} | = 0 \\ | \begin{array}{ccc} x - 2 & y - 2 & z + 1 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{array} | = 0 \end{aligned}$
$\Rightarrow (x - 2) [2 - 3] - (y - 2) [1 - 3] + (z + 1) [1 - 2] = 0$
$\Rightarrow - (x - 2) + 2 (y - 2) - 1 (z + 1) = 0$
$\Rightarrow (x - 2) - 2 (y - 2) + (z + 1) = 0$
$\Rightarrow x - 2 y + z + 3 = 0$
$∴$ Angle between plane $π_{1}$ and $π_{2}$
$\cos θ = | (\frac{0 \times 1 + 1 \times - 2 + 2 \times 1}{\sqrt{0 + 1 + 4} \sqrt{1 + 4 + 1}}) |$
$\cos θ = 0$
$\cos θ = \cos \frac{π}{2}$
$θ = \frac{π}{2}$

TS EAMCET-10.09.2020

Three Dimensional Geometry

121337 A tetrahedron has vertices $O (0, 0, 0), (A (1, 2$ , 1) $B (2, 1, 3), C (- 1, 1, 2)$ . If $θ$ is the angle between the faces $O A B$ and $A B C$ , then $\cos θ =$

1

\frac{1}{\sqrt{2}}

2

\frac{19}{35}

3

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

4

\frac{17}{31}

Explanation:

B Equation of plane OAB is given by,
$| \begin{array}{ccc} x - 0 & y - 0 & z - 0 \\ 1 - 0 & 2 - 0 & 1 - 0 \\ 2 - 0 & 1 - 0 & 3 - 0 \end{array} | = 0 \Rightarrow | \begin{array}{ccc} x & y & z \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array} | = 0$
$x (6 - 1) - y (3 - 2) + z (1 - 4) = 0$
$5 x - y - 3 z = 0$
Equation of plane $ABC$ is given by,
$\begin{aligned} | \begin{array}{ccc} x - 1 & y - 2 & z - 1 \\ 2 - 1 & 1 - 2 & 3 - 1 \\ - 1 - 1 & 1 - 2 & 2 - 1 \end{array} | = 0 \\ | \begin{array}{ccc} x - 1 & y - 2 & z - 1 \\ 1 & - 1 & 2 \\ - 2 & - 1 & 1 \end{array} | = 0 \end{aligned}$
$(x - 1) (- 1 + 2) - (y - 2) (1 + 4) + (z - 1) (- 1 - 2) = 0$
$x - 1 - 5 y + 10 - 3 z + 3 = 0$
$x - 5 y - 3 z + 12 = 0$
Then angle between planes represented by Eqs. (i) and (ii) is given by -
$\cos θ = \frac{(5) (1) + (- 1) (- 5) + (- 3) (- 3)}{\sqrt{25 + 1 + 9} \sqrt{1 + 25 + 9}} = \frac{19}{35}$

TS EAMCET-10.09.2020

Three Dimensional Geometry

121307 If a line makes angles of measure $\frac{π}{6}$ and $\frac{π}{3}$ with $X$ and $Y$ axes respectively, then the angle made by the line with $Z$ axis is

1

\frac{π}{6}

2

\frac{π}{2}

3

\frac{π}{5}

4

\frac{π}{4}

Explanation:

B Let,
| x-axis, | y-axis | z-axis |
| :---: | :---: | :---: |
| $α$ | $β$ | $γ$ |
| $\cos α = 1$ | $\cos β = m$ | $\cos γ = n$ |
We know that,
$l^{2} + m^{2} + n^{2} = 1$
${[\cos (\frac{π}{6})]}^{2} + {[\cos (\frac{π}{3})]}^{2} + (\cos α)^{2} = 1$
$(\frac{3}{4}) + (\frac{1}{4}) + \cos^{2} α = 1$
$1 + \cos^{2} α = 1$
$\cos^{2} α = 0$
$α = \frac{π}{2}$

MHT CET-2020