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7 months ago

Three Dimensional Geometry Class 12th

29. Find the intercepts cut off by the plane $2 x + y - z = 5$

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Vishal Baghel

Contributor-Level 10

$2 x + y - z = 5 . . . . . . . . . (1)$

Dividing both sides of equation (1) by 5, we obtain

$\begin{array}{l} \frac{2}{5} x + \frac{y}{5} - \frac{z}{5} = 1 \\ \Rightarrow \frac{x}{\frac{5}{2}} + \frac{y}{5} + \frac{z}{- 5} = 1 . . . . . . . . . . (2) \end{array}$

It is known that the equation of a plane in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ , where a, b, c are the intercepts cut off by the plane at x, y, and z axes respectively.

Therefore, for the given equation,

$a = \frac{5}{2}, b = 5 a n d c = - 5$

Thus, the intercepts cut off by the plane are $\frac{5}{2}$ $, 5 a n d - 5 .$

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7 months ago

Three Dimensional Geometry Class 12th

28. Find the equations of the planes that passes through three points:

(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)

(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)

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Vishal Baghel

Contributor-Level 10

We know that through three collinear points $A, B, C$ i.e., through a straight line, we can pass an infinite number of planes.

(a) The given points are $A (1, 1, - 1), B (6, 4, - 5), a n d C (- 4, - 2, 3) .$

$\begin{array}{l} | \begin{matrix} 1 & 1 & - 1 \\ 6 & 4 & - 5 \\ - 4 & - 2 & 3 \end{matrix} | = (12 - 10) - (18 - 20) - (- 12 + 16) \end{array}$

$\begin{array}{l} = 2 + 2 - 4 \\ = 0 \end{array}$

Since $A, B, C$ are collinear points, there will be infinite number of planes passing through the given points.

(b) The given points are $A (1, 1, 0), B (1, 2, 1), a n d C (- 2, 2, - 1) .$

$| \begin{matrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ - 2 & 2 & - 1 \end{matrix} | = (- 2 - 2) - (2 + 2) = - 8 \neq 0$

Therefore, a plane will pass through the points A, B, and C.

It is known that the equation of the plane through the points, $(x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) & (x_{3}, y_{3}, z_{3})$ , is

$\begin{array}{l} | \begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} \\ x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ x_{3} - x_{1} & y_{3} - y_{1} & z_{3} - z_{1} \end{matrix} | = 0 \\ \Rightarrow | \begin{matrix} x - 1 & y - 1 & z \\ 0 & 1 & 1 \\ - 3 & 1 & - 1 \end{matrix} | = 0 \\ \Rightarrow (- 2) (x - 1) - 3 (y - 1) + 3 z = 0 \\ \Rightarrow - 2 x - 3 y + 3 z + 2 + 3 = 0 \\ \Rightarrow - 2 x - 3 y + 3 z = - 5 \\ \Rightarrow 2 x + 3 y - 3 z = 5 \end{array}$

This is the Cartesian equation of the required plane.

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

27. Find the vector and Cartesian equations of the planes

(a) That passes through the point $(1, 0, - 2)$ and the normal to the plane is $\hat{i} - \hat{j} - \hat{k}$ .

(b) That passes through the point $(1, 4, 6)$ and the normal vector to the plane is $\hat{i} - 2 \hat{j} + \hat{k}$ .

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Vishal Baghel

Contributor-Level 10

(a) The position vector of point $(1, 0, - 2)$ is $\vec{a} = \hat{i} - 2 \hat{k}$

The normal vector N perpendicular to the plane is $\vec{N} = \hat{i} + \hat{j} - \hat{k}$

The vector equation of the plane is given by, $(\vec{r} - \vec{a}) . \vec{N} = 0$

$\Rightarrow [\vec{r} - (\hat{i} - 2 \hat{k})] . (\hat{i} + \hat{j} - \hat{k}) = 0 . . . . . . . . . (1)$

$\vec{r}$ is the position vector of any point $(x, y, z)$ in the plane.

$∴ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

Therefore, equation (1) becomes

$\begin{array}{l} [(x \hat{i} + y \hat{j} + z \hat{k}) - (\hat{i} - 2 \hat{k})] . (\hat{i} + \hat{j} - \hat{k}) = 0 \\ \Rightarrow [(x - 1) \hat{i} + y \hat{j} + (z + 2) \hat{k}] . (\hat{i} + \hat{j} - \hat{k}) = 0 \\ \Rightarrow (x - 1) + y - (z + 2) = 0 \\ \Rightarrow x + y - z - 3 = 0 \\ \Rightarrow x + y - z = 3 \end{array}$

This is the Cartesian equation of the required plane.

(b) The position vector of the point $(1, 4, 6)$ is $\vec{a} = \hat{i} + 4 \hat{j} + 6 \hat{k}$

The normal vector $\vec{N}$ perpendicular to the plane is $\vec{N} = \hat{i} - 2 \hat{j} + \hat{k}$

The vector equation of the plane is given by, $(\vec{r} - \vec{a}) . \vec{N} = 0$

$\Rightarrow [\vec{r} - (\hat{i} + 4 \hat{j} + 6 \hat{k})] . (\hat{i} - 2 \hat{j} + \hat{k}) = 0 . . . . . . . . . (1)$

$\vec{r}$ is the position vector of any point $P (x, y, z)$ in the plane.

$∴ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

Therefore, equation (1) becomes

$\begin{array}{l} [(x \hat{i} + y \hat{j} + z \hat{k}) - (\hat{i} + 4 \hat{j} + 6 \hat{k})] . (\hat{i} - 2 \hat{j} + \hat{k}) = 0 \\ \Rightarrow [(x - 1) \hat{i} + (y - 4) \hat{j} + (z - 6) \hat{k}] . (\hat{i} - 2 \hat{j} + \hat{k}) = 0 \\ \Rightarrow (x - 1) + 2 (y - 4) + (z - 6) = 0 \\ \Rightarrow x - 2 y + z + 1 = 0 \end{array}$

This is the Car

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(a) The position vector of point $(1, 0, - 2)$ is $\vec{a} = \hat{i} - 2 \hat{k}$

The normal vector N perpendicular to the plane is $\vec{N} = \hat{i} + \hat{j} - \hat{k}$

The vector equation of the plane is given by, $(\vec{r} - \vec{a}) . \vec{N} = 0$

$\Rightarrow [\vec{r} - (\hat{i} - 2 \hat{k})] . (\hat{i} + \hat{j} - \hat{k}) = 0 . . . . . . . . . (1)$

$\vec{r}$ is the position vector of any point $(x, y, z)$ in the plane.

$∴ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

Therefore, equation (1) becomes

This is the Cartesian equation of the required plane.

(b) The position vector of the point $(1, 4, 6)$ is $\vec{a} = \hat{i} + 4 \hat{j} + 6 \hat{k}$

The normal vector $\vec{N}$ perpendicular to the plane is $\vec{N} = \hat{i} - 2 \hat{j} + \hat{k}$

The vector equation of the plane is given by, $(\vec{r} - \vec{a}) . \vec{N} = 0$

$\Rightarrow [\vec{r} - (\hat{i} + 4 \hat{j} + 6 \hat{k})] . (\hat{i} - 2 \hat{j} + \hat{k}) = 0 . . . . . . . . . (1)$

$\vec{r}$ is the position vector of any point $P (x, y, z)$ in the plane.

$∴ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

Therefore, equation (1) becomes

This is the Cartesian equation of the required plane.

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

26. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin:

$\begin{array}{l} \end{array} (a) 2 x + 3 y + 4 z - 12 = 0$

$(b) 3 y + 4 z - 6 = 0$

$(c) x + y + z = 1$

$(d) 5 y + 8 = 0$

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Vishal Baghel

Contributor-Level 10

(a) Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1}, y_{1}, z_{1}) .$

$\begin{array}{l} 2 x + 3 y + 4 z - 12 = 0 \end{array}$

$\begin{array}{l} \Rightarrow 2 x + 3 y + 4 z = 12 \dots (1) \end{array}$

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

25. Find the Cartesian equation of the following planes:

$\begin{array}{l} (a) \vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2 \\ (b) \vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1 \\ (c) \vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 \end{array}$

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Vishal Baghel

Contributor-Level 10

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value

...more

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

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

$\begin{array}{l} (x \hat{i} + y \hat{j} - z \hat{k}) . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 \\ \Rightarrow (s - 2 t) x + (3 - t) y + (2 s + t) z = 15 \end{array}$

This is the Cartesian equation of the given plane.

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7 months ago

Three Dimensional Geometry Class 12th

24. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3 \hat{i} + 5 \hat{j} - 6 \hat{k}$

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Vishal Baghel

Contributor-Level 10

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value

...more

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

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

$\begin{array}{l} (x \hat{i} + y \hat{j} - z \hat{k}) . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 \\ \Rightarrow (s - 2 t) x + (3 - t) y + (2 s + t) z = 15 \end{array}$

This is the Cartesian equation of the given plane.

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

23. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

$(a$ $)$ $z = 2$

$(b) x + y + z = 1$

$(c) 2 x + 3 y - z = 5$

$(d) 5 y + 8 = 0$

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Vishal Baghel

Contributor-Level 10

(a) The equation of the plane is $z = 2 o r 0 x + 0 y + z = 2 . . . . . . . . . . (1)$

The direction ratios of normal are $0, 0, a n d 1 .$

$∴ + 02 + 12 = 1$

Dividing both sides of equation (1) by 1, we obtain

$0 . x + 0 . y + 1 . z = 2$

This is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) $x + y + z = 1 . . . . . . . . . . (1)$

The direction ratios of normal are 1, 1, and 1.

$∴ + (1) ² + (1) ² =$

Dividing both sides of equation (1) by , we obtain

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{} . . . . . . . . . (2)$

This equation is of the form $l x + m y + n z = d$ , where l, m, n

...more

(a) The equation of the plane is $z = 2 o r 0 x + 0 y + z = 2 . . . . . . . . . . (1)$

The direction ratios of normal are $0, 0, a n d 1 .$

$∴ + 02 + 12 = 1$

Dividing both sides of equation (1) by 1, we obtain

$0 . x + 0 . y + 1 . z = 2$

This is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) $x + y + z = 1 . . . . . . . . . . (1)$

The direction ratios of normal are 1, 1, and 1.

$∴ + (1) ² + (1) ² =$

Dividing both sides of equation (1) by , we obtain

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{} . . . . . . . . . (2)$

This equation is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal are $\frac{1}{}$ , $\frac{1}{}$ and $\frac{1}{}$ the distance of normal from the origin is $\frac{1}{}$ units.

(c) $2 x + 3 y - z = 5 . . . . . . . . . . (1)$

The direction ratios of normal are 2, 3, and −1.

$∴ + (3) ² + (- 1) ² =$

Dividing both sides of equation (1) by , we obtain

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

This equation is of the form $l x + m y + n z = d$ where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane $\frac{2}{x} + \frac{3}{y} - \frac{1}{z}$ and the distance of normal from the origin is $\frac{5}{}$ units.

(d)

$\begin{array}{l} 5 y + 8 = 0 & \Rightarrow 0 x - 5 y + 0 z = 8 . . . . . . . . . . (1) \end{array}$

The direction ratios of normal are 0, −5, and 0.

$∴ + (- 5) ² + 0 = 5$

Dividing both sides of equation (1) by 5, we obtain

$- y = \frac{8}{5}$

This equation is of the form $l x + m y + n z = d$ where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the distance of normal from the origin is $\frac{8}{5}$ units.

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

22. Find the shortest distance between the lines whose vector equations are

$\begin{array}{l} \vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} = (3 - 2 t) \hat{k} a n d \\ \vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} = (2 s + 1) \hat{k} \end{array}$

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Vishal Baghel

Contributor-Level 10

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

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

Here,

Hence, the shortage distance between the line is given by

$d = | \frac{({\vec{b}}_{1} * {\vec{b}}_{2}) . ({\vec{a}}_{2} - {\vec{a}}_{1})}{| {\vec{b}}_{1} * {\vec{b}}_{2} |} | = 8$

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

21. Find the shortest distance between the lines whose vector equations are

$\begin{array}{l} \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} + 3 \hat{j} + 2 \hat{k}) \\ a n d \vec{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + \hat{k}) \end{array}$

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - 3 \hat{j} + 2 \hat{k}) a n d - - - - - (1) \\ \vec{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + \hat{k}) - - - - - (2) \end{array}$

Here, comparing (1) and (2) with $\vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ and $\vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ , we have

$\begin{array}{l} a_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}, b_{1} = \hat{i} - 3 \hat{j} + 2 \hat{k} \\ a_{2} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k}, b_{2} = 2 \hat{i} + 3 \hat{j} + \hat{k} \end{array}$

Therefore,

\begin{array}{l} {\vec{a}}_{2} - {\vec{a}}_{1} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \\ {\vec{b}}_{1} * {\vec{b}}_{2} = \\ = \hat{i} (- 3 - 6) - \hat{j} (1 - 4) + \hat{k} (3 + 6) \\ = - 9 \hat{i} + 3 \hat{j} + 9 \hat{k} \\ | {\vec{b}}_{1} * {\vec{b}}_{2} | = = \\ = \\ = 3 \end{array}

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

20. Find the shortest distance between the lines $\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} a n d \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$ .

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \\ \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \end{array}$

Shortest distance between two lines is given by,

given equation we have

$\begin{array}{l} x_{1} = - 1, y_{1} = - 1, z_{1} = - 1 \\ a_{1} = 7, b_{1} = - 6, c_{1} = 1 \\ x_{2} = 3, y_{2} = 5, z_{2} = 7 \\ a_{2} = 1, b_{2} = - 2, c_{2} = 1 \end{array}$

Then,

$\begin{array}{l} = 4 (- 6 + 2) - 6 (7 - 1) + 8 (- 14 + 6) \\ = - 16 - 36 - 64 \\ = - 116 \end{array}$

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