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6 months ago

Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1} a n d \frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3},$ is

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Payal Gupta

Contributor-Level 10

$L_{1} : \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1}$

$L_{2} : \frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3}$

Now,

$\vec{p} * \vec{q} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & - 1 \\ 2 & 1 & 3 \end{matrix} | = 10 \hat{i} - 8 \hat{j} - 4 \hat{k}$

and

${\vec{a}}_{2} - {\vec{a}}_{1} = 6 \hat{i} - 4 \hat{j} - 4 \hat{k}$

$∴ S . D = | \frac{60 + 32 + 16}{\sqrt[]{100 + 64 + 16}} | = \frac{108}{\sqrt[]{180}} = \frac{18}{\sqrt[]{5}}$

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6 months ago

Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let the foot of the perpendicular from the point (1, 2, 4) on the line $\frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3}$ be P. Then the distance of P from the plane 3x + 4y + 12z + 23 = 0 is

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Payal Gupta

Contributor-Level 10

$L_{1} : \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1}$

$L_{2} : \frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3}$

Now,

$\vec{p} * \vec{q} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & - 1 \\ 2 & 1 & 3 \end{matrix} | = 10 \hat{i} - 8 \hat{j} - 4 \hat{k}$

and

${\vec{a}}_{2} - {\vec{a}}_{1} = 6 \hat{i} - 4 \hat{j} - 4 \hat{k}$

$∴ S . D = | \frac{60 + 32 + 16}{\sqrt[]{100 + 64 + 16}} | = \frac{108}{\sqrt[]{180}} = \frac{18}{\sqrt[]{5}}$

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

59. The planes: $2 x - y + 4 z = 5 a n d 5 x - 2.5 y + 10 z = 6$ are

(A) Perpendicular (B) Parallel (C) intersect y-axis

(D) Passes through (0,0,5/4)

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Vishal Baghel

Contributor-Level 10

The equations of the planes are

$\begin{array}{l} 2 x - y + 4 z = 5 \dots (1) & 5 x - 2.5 y + 10 z = 6 \dots (2) \end{array}$

It can be seen that,

$\begin{array}{l} \frac{a_{1}}{a_{2}} = \frac{2}{5} \\ \frac{b_{1}}{b_{2}} = \frac{- 1}{- 2.5} = \frac{2}{5} \\ \frac{c_{1}}{c_{2}} = \frac{4}{10} = \frac{2}{5} \\ ∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{array}$

Therefore, the given planes are parallel.

Hence, the correct answer is B.

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7 months ago

Three Dimensional Geometry Class 12th

Choose the correct answer in Exercise Q. 22 and 23.

58. Distance between the two planes: $2 x + 3 y + 4 z = 4 a n d 4 x + 6 y + 8 z = 12$ is

(A) 2 units (B) 4 units (C) 8 units (D) 2/√29 units

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Vishal Baghel

Contributor-Level 10

The equations of the planes are

$\begin{array}{l} 2 x + 3 y + 4 z = 4 & 4 x + 6 y + 8 z = 12 & \Rightarrow 2 x + 3 y + 4 z = 6 \end{array}$

It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes, $a x + b y + c z = d_{1} a n d a x + b y + c z = d_{2},$ is given by,

$\begin{array}{l} D = | \frac{d_{2} - d_{1}}{} | \\ \Rightarrow D = | \frac{6 - 4}{} | \\ D = \frac{2}{} \end{array}$

Thus, the distance between the lines is 2/√29 units.

Hence, the correct answer is D.

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

57. Prove that if a plane has the intercepts a,b,c and is at a distance of P units from the origin, then $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$

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Vishal Baghel

Contributor-Level 10

The equation of a plane having intercepts a, b, c with x, y, and z axes respectively is given by,

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

The distance (p) of the plane from the origin is given by,

$\begin{array}{l} p = | \frac{\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1}{} | \\ \Rightarrow p = \frac{1}{} \\ \Rightarrow p^{2} = \frac{1}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} \\ \Rightarrow \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} \end{array}$

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

56. Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines:

$\frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}$

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Vishal Baghel

Contributor-Level 10

Let the required line be parallel to the vector $\vec{b}$ given by, $\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

The position vector of the point (1, 2, − 4) is $\vec{a} = \hat{i} + 2 \hat{j} - 4 \hat{k}$

The equation of the line passing through (1, 2, −4) and parallel to vector $\vec{b}$ is

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \Rightarrow \vec{r} = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . . . . . . . (1) \end{array}$

The equations of the lines are

$\begin{array}{l} \frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7} . . . . . . . . (2) \\ \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5} . . . . . . . . (3) \end{array}$

Line (1) and line (2) are perpendicular to each other.

$∴ 3 b_{1} - 16 b_{2} + 7 b_{3} = 0 . . . . . . . . (4)$

Also, line (1) and line (3) are perpendicular to each other.

$∴ 3 b_{1} + 8 b_{2} - 5 b_{3} = 0 . . . . . . . . (5)$

From equations (4) and (5), we obtain

$\begin{array}{l} \frac{b_{1}}{(- 16) (- 5) - 8 * 7} = \frac{b_{2}}{7 * 3 - 3 (- 5)} = \frac{b_{3}}{3 * 8 - 3 (- 16)} \\ \Rightarrow \frac{b_{1}}{24} = \frac{b_{2}}{36} = \frac{b_{3}}{72} \\ \Rightarrow \frac{b_{1}}{2} = \frac{b_{2}}{3} = \frac{b_{3}}{6} \end{array}$

Direction ratios of $\vec{b}$ are 2, 3, and 6.

$∴ \vec{b} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$

Substituting $\vec{b} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ in equation (1), we obtain

$\vec{r} = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$

This is the equation of the required line.

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

55. Find the vector equation of the line passing through (1, 2, 3) and parallel to the plane $\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5$ and $\vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6$

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Vishal Baghel

Contributor-Level 10

Let the required line be parallel to vector $\vec{b}$ given by,

$\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

The position vector of the point (1, 2, 3) is $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

The equation of line passing through (1, 2, 3) and parallel to $\vec{b}$ is given by,

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \Rightarrow \vec{r} (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . . . . . . . . (1) \end{array}$

The equations of the given planes are

$\begin{array}{l} \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5 . . . . . . . . (2) \\ \vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6 . . . . . . . . (3) \end{array}$

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

$\begin{array}{l} \Rightarrow (\hat{i} - \hat{j} + 2 \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0 \\ \Rightarrow λ (b_{1} - b_{2} + 2 b_{3}) = 0 \\ \Rightarrow b_{1} - b_{2} + 2 b_{3} = 0 . . . . . . . . . . (4) \end{array}$

Similarly, $(3 \hat{i} + \hat{j} + \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0$

From equations (4) and (5), we obtain

$\begin{array}{l} \frac{b_{1}}{(- 1) * 1 - 1 * 2} = \frac{b_{2}}{2 * 3 - 1 * 1} = \frac{b_{3}}{1 * 1 - 3 * (- 1)} \\ \Rightarrow \frac{b_{1}}{- 3} = \frac{b_{2}}{5} = \frac{b_{3}}{4} \end{array}$

Therefore, the direction ratios of $\vec{b}$ are −3, 5, and 4.

$∴ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = - 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$

Substituting the value of $\vec{b}$ in equation (1), we obtain

$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 3 \hat{i} + 5 \hat{j} + 4 \hat{k})$

This is the equat

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Let the required line be parallel to vector $\vec{b}$ given by,

$\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

The position vector of the point (1, 2, 3) is $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

The equation of line passing through (1, 2, 3) and parallel to $\vec{b}$ is given by,

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \Rightarrow \vec{r} (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . . . . . . . . (1) \end{array}$

The equations of the given planes are

$\begin{array}{l} \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5 . . . . . . . . (2) \\ \vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6 . . . . . . . . (3) \end{array}$

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

Similarly, $(3 \hat{i} + \hat{j} + \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0$

From equations (4) and (5), we obtain

$\begin{array}{l} \frac{b_{1}}{(- 1) * 1 - 1 * 2} = \frac{b_{2}}{2 * 3 - 1 * 1} = \frac{b_{3}}{1 * 1 - 3 * (- 1)} \\ \Rightarrow \frac{b_{1}}{- 3} = \frac{b_{2}}{5} = \frac{b_{3}}{4} \end{array}$

Therefore, the direction ratios of $\vec{b}$ are −3, 5, and 4.

$∴ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = - 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$

Substituting the value of $\vec{b}$ in equation (1), we obtain

$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 3 \hat{i} + 5 \hat{j} + 4 \hat{k})$

This is the equation of the required line.

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

53. Find the equation of the plane which contains the line of intersection of the planes $\vec{r} (\hat{i} + 2 \hat{j} + 3 \hat{k}) - 4 = 0, \vec{r} . (2 \hat{i} + h t \hat{j} - \hat{k}) + 5 = 0$ and which is perpendicular to the plane $\vec{r} . (5 \hat{i} + 3 \hat{j} - 6 \hat{k}) + 8 = 0$

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Vishal Baghel

Contributor-Level 10

The equations of the given planes are

$\begin{array}{l} \vec{r} . (\hat{i} + 2 \hat{j} + 3 \hat{k}) - 4 = 0 . . . . . (1) \\ \vec{r} . (2 \hat{i} + \hat{j} - \hat{k}) + 5 = 0 . . . . . . (2) \end{array}$

The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is

$\begin{array}{l} [\vec{r} . (\hat{i} + 2 \hat{j} + 3 \hat{k}) - 4] + λ [\vec{r} . (2 \hat{i} + \hat{j} - \hat{k}) + 5] = 0 \\ \vec{r} . [(2 λ + 1) \hat{i} + (λ + 2) \hat{j} + (3 - λ) \hat{k}] + (5 λ - 4) = 0 . . . . . (3) \end{array}$

The plane in equation (3) is perpendicular to the plane, $\vec{r} . (5 \hat{i} + 3 \hat{j} - 6 \hat{k}) + 8 = 0$

$\begin{array}{l} ∴ 5 (2 λ + 1) + 3 (λ + 2) - 6 (3 - λ) = 0 \\ \Rightarrow 19 λ - 7 = 0 \\ \Rightarrow λ = \frac{7}{19} \end{array}$

Substituting λ = 7/19 in equation (3), we obtain

$\begin{array}{l} \Rightarrow \vec{r} . (\frac{13}{19} \hat{i} + \frac{45}{19} \hat{j} + \frac{50}{19} \hat{k}) \frac{- 41}{19} = 0 \\ \Rightarrow \vec{r} . (33 \hat{i} + 45 \hat{j} + 50 \hat{k}) - 41 = 0 \end{array}$

This is the vector equation of the required plane.

The Cartesian equation of this plane can be obtained by substituting $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ in equation (3).

$\begin{array}{l} (x \hat{i} + y \hat{j} + z \hat{k}) . (33 \hat{i} + 45 \hat{j} + 50 \hat{k}) - 41 = 0 \\ \Rightarrow 33 x + 45 y + 50 z - 41 = 0 \end{array}$

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

52. If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the plane passing through P and perpendicular to OP.

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Vishal Baghel

Contributor-Level 10

The coordinates of the points, O and P, are (0, 0) and (1, 2, −3) respectively.

Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3

It is known that the equation of the plane passing through the point (x1, y1 z1) is

$a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$ where, a, b, and c are the direction ratios of normal.

Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).

Thus, the equation of the required plane is

$\begin{array}{l} \end{array}$ $\begin{array}{l} 1 (x - 1) + 2 (y - 2) - 3 (z + 3) 0 & \Rightarrow x + 2 y - 3 z - 14 = 0 \end{array}$

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

51. Find the equation of the plane passing through the line of intersection of the planes $\vec{r} . (\hat{i} + \hat{j} + \hat{k}) = 1 & \vec{r} . (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0$ and parallel to x - axis.

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Vishal Baghel

Contributor-Level 10

Equation of one plane is

$\begin{array}{l} \vec{r} . (\hat{i} + \hat{j} + \hat{k}) = 1 \\ \Rightarrow \vec{r} . (\hat{i} + \hat{j} + \hat{k}) - 1 = 0 \\ \vec{r} . (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0 \end{array}$

The equation of any plane passing through the line of intersection of these planes is

$\begin{array}{l} [\vec{r} . (\hat{i} + \hat{j} + \hat{k}) - 1] + λ [\vec{r} . (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4] = 0 \\ \vec{r} . [(2 λ + 1) \hat{i} + (3 λ + 1) \hat{j} + (1 - λ) \hat{k}] + (4 λ - 1) = 0 . . . . . (1) \end{array}$

Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-axis are 1, 0, and 0.

$\begin{array}{l} ∴ 1 . (2 λ + 1) + 0 (3 λ + 1) + 0 (1 - λ) = 0 \\ \Rightarrow 2 λ + 1 = 0 \\ \Rightarrow λ = - \frac{1}{2} \end{array}$

Substituting λ = -1/2 in equation (1), we obtain

$\begin{array}{l} \Rightarrow \vec{r} . [- \frac{1}{2} \hat{j} + \frac{3}{2} \hat{k}] + (- 3) = 0 \\ \Rightarrow \vec{r} (\hat{j} - 3 \hat{k}) + 6 = 0 \end{array}$

Therefore, its Cartesian equation is y − 3z + 6 = 0

This is the equation of the required plane.

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