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Three Dimensional Geometry Class 12th

40. Find the equation of the line parallel to x-axis and passing through the origin.

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Vishal Baghel

Contributor-Level 10

The line parallel to x-axis and passing through the origin is x-axis itself.

Let A be a point on x-axis. Therefore, the coordinates of A are given by $(a, 0, 0), w h e r e a ? R .$ Direction ratios of $O A a r e (a ? 0) = a, 0, 0$

The equation of OA is given by,

$\begin{array}{l} \frac{x ? 0}{a} = \frac{y ? 0}{0} = \frac{z ? 0}{0} \\ ? \frac{x}{1} = \frac{y}{0} = \frac{z}{0} = a \end{array}$

Thus, the equation of line parallel to x-axis and passing through origin is

$\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$

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Three Dimensional Geometry Class 12th

39. If $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are $m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - l_{2} m_{1} .$

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Vishal Baghel

Contributor-Level 10

It is given that $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines. Therefore,

$\begin{array}{l} l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0 . . . . . . . . . . (1) \\ {l_{1}}^{2} + {m_{1}}^{2} + n_{1}^{2} = 1 . . . . . . . . . . (2) \\ {l_{2}}^{2} + {m_{2}}^{2} + n_{2}^{2} = 1 . . . . . . . . . . (3) \end{array}$

Let $l, m, n$ be the direction cosines of the line which is perpendicular to the line with direction cosines $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2} .$

$\begin{array}{l} ∴ l l_{1} + m m_{1} + n n_{1} = 0 \\ l l_{2} + m m_{2} + n n_{2} = 0 \\ ∴ \frac{l}{m_{1} n_{2} - m_{2} n_{1}} = \frac{m}{n_{1} l_{2} - n_{2} l_{1}} = \frac{n}{l_{1} m_{2} - l_{2} m_{1}} \\ \Rightarrow \frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{1})}^{2}} \\ \Rightarrow \frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{2})}^{2}} \\ = \frac{l^{2} + m^{2} + n^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{2})}^{2}} . . . . . . . . . . (4) \end{array}$

$l, m, n$ are the direction cosines of the line.

$∴ l^{2} + m^{2} + n^{2} = 1 \dots (5)$

It is known that,

$\begin{array}{l} ({l_{1}}^{2} + {m_{1}}^{2} + n_{1}^{2}) ({l_{2}}^{2} + {m_{2}}^{2} + n_{2}^{2}) - {(l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})}^{2} \\ = {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} \\ F r o m, (1), (2) & (3), w e . o b t a i n \\ \Rightarrow 1.1 - 0 = {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} \\ ∴ {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} = 1 . . . . . . . . . . (6) \end{array}$

Substituting the values from equations (5) and (6) in equation (4), we obtain

$\frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{1})}^{2}} = 1$

Thus, the direction cosines of the required line are $m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - l_{2} m_{1}$

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7 months ago

Three Dimensional Geometry Class 12th

38. Show that the line joining the origin to the point $(2, 1, 1)$ is perpendicular to the line determined by the points $(3, 5, - 1), (4, 3, - 1) .$

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Vishal Baghel

Contributor-Level 10

Let OA be the line joining the origin, $O (0, 0, 0),$ and the point, $A (2, 1, 1) .$

Also, let BC be the line joining the points, $B (3, 5, - 1) a n d C (4, 3, - 1) .$

The direction ratios of $O A a r e 2, 1, a n d 1 a n d o f B C a r e (4 - 3) = 1, (3 - 5) = - 2, a n d (- 1 + 1) = 0$

OA is perpendicular to $B C, i f a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$∴ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 1 + 1 (- 2) + 1 * 0 = 2 - 2 = 0$

Thus, OA is perpendicular to BC.

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7 months ago

Three Dimensional Geometry Class 12th

37. In the following cases find the distances of each of the given points from the corresponding given plane:

$\begin{array}{l} \end{array} (a) P o i n t (0, 0, 0) P l a n e 3 x - 4 y + 12 z = 3$

$(b) P o i n t (3, - 2, 1) P l a n e 2 x - y + 2 z + 3 = 0$

$c) P o i n t (2, 3, - 5) P l a n e x + 2 y - 2 z = 9$

$(d) P o i n t (- 6, 0, 0) P l a n e 2 x - 3 y + 6 z - 2 = 0$

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Vishal Baghel

Contributor-Level 10

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Three Dimensional Geometry Class 12th

36. In the following cases, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angle between them.

$\begin{array}{l} \end{array} (a) 7 x + 5 y + 6 z + 30 = 0 a n d 3 x - y - 10 z + 4 = 0$

$(b) 2 x + y + 3 z - 2 = 0 a n d x - 2 y + 5 = 0$

$(c) 2 x - 2 y + 4 z + 5 = 0 a n d 3 x - 3 y + 6 z - 1$

$(d) 2 x - y + 3 z - 1 = 0 a n d 2 x - y + 3 z + 3 = 0$

$(e) 4 x + 8 y + z - 8 = 0 a n d y + z - 4 = 0$

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Vishal Baghel

Contributor-Level 10

The direction ratios of normal to the plane, $L_{1} : a_{1} x + b_{1} y + c_{1} z = 0 .$ , are $a_{1}, b_{1}, c_{1}$ and $L_{2} : a_{2} x + b_{2} y + c_{2} z = 0 .$

$L_{1}$
$\begin{array}{l} L_{2}, i f \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ L_{1} ⊥ L_{2}, i f a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \end{array}$

The angle between $L_{1} & L_{2}$ is given by,

(b) The equations of the planes are $2 x + y + 3 z - 2 = 0 a n d x - 2 y + 5 = 0$

$\begin{array}{l} a_{1} = 2, b_{1} = 1, c_{1} = 3 & a_{2} =, b_{2} = - 2, c_{2} = \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 1 + 1 * (- 2) + 3 * 0 = 0 \end{array}$

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are $2 x - 2 y + 4 z + 5 = 0 a n d 3 x - 3 y + 6 z - 1$

Here, $a_{1} = 2, b_{1} = - 2, c_{1} = 4 & a_{2} =, b_{2} = - 3, c_{2} = 6$

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 3 + 1 * (- 2) (- 3) + 4 * 6 = 6 + 6 + 24 = 36 \neq 0$

Thus, the given planes are not perpendicular to each other.

$\begin{array}{l} \frac{a_{1}}{a_{2}} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{- 2}{- 3} = \frac{2}{3} & \frac{c_{1}}{c_{2}} = \frac{4}{6} = \frac{2}{3} \\ ∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{array}$

Thus, the given planes are parallel to each other

(d) The equations of the planes are and $2 x - y + 3 z - 1 = 0 a n d 2 x - y + 3 z + 3 = 0$

$\begin{array}{l} a_{1} = 2, b_{1} = - 1, c_{1} = 3 & a_{2} =, b_{2} = - 1, c_{2} = 3 \\ \frac{a_{1}}{a_{2}} = \frac{2}{2} = 1, \frac{b_{1}}{b_{2}} = \frac{- 1}{- 1} = 1 & \frac{c_{1}}{c_{2}} = \frac{3}{3} = 1 \\ ∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{array}$

Thus, the given lines are parallel to each other

(e) The equations of the given planes are $4 x + 8 y + z - 8 = 0 a n d y + z - 4 = 0$ $\begin{array}{l} a_{1} = 4, b_{1} = 8, c_{1} = 1 & a_{2} =, b_{2} = 1, c_{2} = 1 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 4 * 0 + 8 * 1 + 1 = 9 \neq 0 \end{array}$

Therefore, the given lines are not perpendicular to each

...more

The direction ratios of normal to the plane, $L_{1} : a_{1} x + b_{1} y + c_{1} z = 0 .$ , are $a_{1}, b_{1}, c_{1}$ and $L_{2} : a_{2} x + b_{2} y + c_{2} z = 0 .$

$L_{1}$
$\begin{array}{l} L_{2}, i f \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ L_{1} ⊥ L_{2}, i f a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \end{array}$

The angle between $L_{1} & L_{2}$ is given by,

(b) The equations of the planes are $2 x + y + 3 z - 2 = 0 a n d x - 2 y + 5 = 0$

$\begin{array}{l} a_{1} = 2, b_{1} = 1, c_{1} = 3 & a_{2} =, b_{2} = - 2, c_{2} = \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 1 + 1 * (- 2) + 3 * 0 = 0 \end{array}$

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are $2 x - 2 y + 4 z + 5 = 0 a n d 3 x - 3 y + 6 z - 1$

Here, $a_{1} = 2, b_{1} = - 2, c_{1} = 4 & a_{2} =, b_{2} = - 3, c_{2} = 6$

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 3 + 1 * (- 2) (- 3) + 4 * 6 = 6 + 6 + 24 = 36 \neq 0$

Thus, the given planes are not perpendicular to each other.

Thus, the given planes are parallel to each other

(d) The equations of the planes are and $2 x - y + 3 z - 1 = 0 a n d 2 x - y + 3 z + 3 = 0$

Thus, the given lines are parallel to each other

Therefore, the given lines are not perpendicular to each other.

$\begin{array}{l} \frac{a_{1}}{a_{2}} = \frac{4}{0}, \frac{b_{1}}{b_{2}} = \frac{8}{1} = 8 & \frac{c_{1}}{c_{2}} = \frac{1}{1} = 1 \\ ∴ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \end{array}$

Therefore, the given lines are not parallel to each other.

The angle between the planes is given by,

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

35. Find the angle between the planes whose vector equations are $\vec{r} . (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5 & \vec{r} . (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3$

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Vishal Baghel

Contributor-Level 10

The equations of the given planes are $\vec{r} . (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5 & \vec{r} . (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3$

It is known that if n1 and n2 are normal to the planes, $\vec{r} . \vec{n_{1}} = d_{1} & \vec{r} . \vec{n_{2}} = d_{2}$ then the angle between them, Q, is given by,

$c o s Q = | \frac{\vec{n_{1}} . \vec{n_{2}}}{| \vec{n_{1}} | | \vec{n_{2}} |} | . . . . . . . . . . (1)$

$\begin{array}{l} H e r e, \vec{n_{1}} = 2 \hat{i} + 2 \hat{j} - 3 \hat{k} & \vec{n_{2}} = 3 \hat{i} - 3 \hat{j} + 5 \hat{k} \\ ∴ \vec{n_{1}} . \vec{n_{2}} = (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 2.3 + 2 . (- 3) + (- 3) . 5 = - 15 \\ | \vec{n_{1}} | = = \\ | \vec{n_{2}} | = = \end{array}$

Substituting the value of $\vec{n_{1}} . \vec{n_{2}}$ $| \vec{n_{1}} | & | \vec{n_{2}} |$ in equation (1), we obtain

$\begin{array}{l} c o s Q = | \frac{- 15}{.} | \\ \Rightarrow c o s Q = \frac{15}{} \\ \Rightarrow c o s Q^{- 1} = (\frac{15}{}) \end{array}$

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New answer posted

7 months ago

Three Dimensional Geometry Class 12th

34. Find the equation of the plane through the line of intersection of the planes $x + y + z = 1 a n d 2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$

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Vishal Baghel

Contributor-Level 10

The equation of the plane through the intersection of the planes, $x + y + z = 1 a n d 2 x + 3 y + 4 z = 5$ , is

$(x + y + z - 1) + λ (2 x + 3 y + 4 z - 5)$

$\Rightarrow (2 λ + 1) x + (3 λ + 1) y + (4 λ + 1) z - (5 λ + 1) = 0 . . . . . . . . . . (1)$

The direction ratios, $a_{1}, b_{1}, c_{1},$ of this plane are $(2 λ + 1), (3 λ + 1), a n d (4 λ + 1) .$

The plane in equation (1) is perpendicular to $x - y + z = 0$

Its direction ratios, $a_{2}, b_{2}, c_{2},$ are $1, - 1, a n d 1$ .

Since the planes are perpendicular,

$\begin{array}{l} a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \\ \Rightarrow (2 λ + 1) - (3 λ + 1) + (4 λ + 1) = 0 \\ \Rightarrow 3 λ + 1 = 0 \\ \Rightarrow λ = - \frac{1}{3} \end{array}$

Substituting $λ = - \frac{1}{3}$ in equation (1), we obtain

$\begin{array}{l} \frac{1}{3} x - \frac{1}{3} z + \frac{2}{3} = 0 \\ \Rightarrow x - z + 2 = 0 \end{array}$

This is the required equation of the plane.

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7 months ago

Three Dimensional Geometry Class 12th

33. Find the vector equation of the plane passing through the intersection of the planes $\vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) = 7, \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$ and through the point $(2, 1, 3) .$

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Vishal Baghel

Contributor-Level 10

The equations of the planes are $\vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) = 7, \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$

$\begin{array}{l} \Rightarrow \vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) - 7 = 0 . . . . . . . . . . (1) \\ \Rightarrow \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) - 9 = 0 . . . . . . . . . . (2) \end{array}$

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

$\begin{array}{l} [\vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) - 7] + λ [\vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) - 9] = 0, w h e r e, λ \in R \\ \vec{r} . [(2 \hat{i} + 2 \hat{j} - 2 \hat{k}) + λ (2 \hat{i} + 5 \hat{j} + 3 \hat{k})] = 9 λ + 7 \\ \vec{r} . [(2 + 2 λ) \hat{i} + (2 + 5 λ) \hat{j} + (3 λ - 3) \hat{k}] = 9 λ + 7 . . . . . . . . . . (3) \end{array}$

The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,

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

Substituting in equation (3), we obtain

$\begin{array}{l} (2 \hat{i} + \hat{j} + 3 \hat{k}) . [(2 + 2 λ) \hat{i} + (2 + 5 λ) \hat{j} + (3 λ - 3) \hat{k}] = 9 λ + 7 \\ \Rightarrow 2 (2 + 2 λ) + 1 (2 + 5 λ) + 3 (3 λ - 3) = 9 λ + 7 \\ \Rightarrow 4 + 4 λ + 2 + 5 λ + 9 λ - 9 = 9 λ + 7 \\ \Rightarrow 18 λ - 3 = 9 λ + 7 \\ \Rightarrow 9 λ = 10 \\ \Rightarrow λ = \frac{10}{9} \end{array}$

Substituting $λ = \frac{10}{9}$ in equation (3), we obtain

$\begin{array}{l} \vec{r} . (\frac{38}{9} \hat{i} + \frac{68}{9} \hat{j} + \frac{3}{9} \hat{k}) = 17 \\ \Rightarrow \vec{r} . (38 \hat{i} + 68 \hat{j} + 3 \hat{k}) = 153 \end{array}$

This is the vector equation of the required plane.

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Three Dimensional Geometry Class 12th

31. Find the equation of the plane through the intersection of the planes $3 x - y + 2 z - 4 = 0 a n d x +$ $y + z - 2 = 0$ and the point $(2, 2, 1) .$

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Vishal Baghel

Contributor-Level 10

The equation of any plane through the intersection of the planes,

$3 x - y + 2 z - 4 = 0 a n d x + y + z - 2 = 0,$ is

$(3 x - y + 2 z - 4) + α (x + y + z - 2) = 0, w h e r e, α \in R . . . . . . . . . . (1)$

The plane passes through the point $(2, 2, 1) .$ Therefore, this point will satisfy equation (1).

$\begin{array}{l} ∴ (3 * 2 - 2 + 2 * 1 - 4) + α (2 + 2 + 1 - 2) = 0 \\ \Rightarrow 2 + 3 α = 0 \\ \Rightarrow α = - \frac{2}{3} \end{array}$

Substituting $α = - \frac{2}{3}$ in equation (1), we obtain

$\begin{array}{l} (3 x - y + 2 z - 4) - \frac{2}{3} (x + y + z - 2) = 0 \\ \Rightarrow 3 (3 x - y + 2 z - 4) - 2 (x + y + z - 2) = 0 \\ \Rightarrow (9 x - 3 y + 6 z - 12) - 2 (x + y + z - 2) = 0 \\ \Rightarrow 7 x - 5 y + 4 z - 8 = 0 \end{array}$

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New answer posted

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Three Dimensional Geometry Class 12th

30. Find the equation of the plane with intercept 3 on the y- axis and parallel to ZOX plane.

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Vishal Baghel

Contributor-Level 10

The equation of the plane ZOX is

y = 0

Any plane parallel to it is of the form, y = a

Since the y-intercept of the plane is 3,

∴ a = 3

Thus, the equation of the required plane is y = 3

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