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Maths Three Dimensional Geometry

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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New answer posted

6 months ago

Maths Three Dimensional Geometry

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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New answer posted

6 months ago

Maths Three Dimensional Geometry

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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New answer posted

6 months ago

Maths Trigonometric Functions

27. sin (n + 1)xsin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

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Payal Gupta

Contributor-Level 10

27. L.H.S $L.HS = s i n (x + 1) x s i n (x + 2) x + c o s (x + 1) x c o s (x + 2) x .$

$= c o s (x + 1) x c o s (x + 2) x + s i n (x + 1) x s i n (x + 2) x .$

Let A = (n+1)x and B = (n+2)x

So, L.H.S = cosAcosB + sin Asin B

$= c o s (A - B)$

Putting values of A and B we get,

L.H.S = $L.H.S = c o s [(n + 1) x - (n + 2) x]$

$= c o s [n x + x - n x - 2 x]$

$= c o s (- x)$

$= c o s x = R.H.S$ R.H.S

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New answer posted

6 months ago

Maths Three Dimensional Geometry

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

6 months ago

Maths Three Dimensional Geometry

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

6 months ago

Maths Three Dimensional Geometry

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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New answer posted

6 months ago

Maths Three Dimensional Geometry

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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New question posted

6 months ago

Maths Trigonometric Functions

27. sin (n + 1)xsin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

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New answer posted

6 months ago

Maths Trigonometric Functions

26. $c o s (\frac{3 π}{2} + x) c o s (2 π + x) [c o t (\frac{3 π}{2} - x) + c o t (2 π + x)] = 1$

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Payal Gupta

Contributor-Level 10

26. L.H.S $L.H.S. = c o s (\frac{3 π}{2} + x) c o s (2 π + x) [c o t (\frac{3 π}{2} - x) + c o t (2 π + x)]$

Here,

$c o s (\frac{3 π}{2} + x) = c o s (\frac{4 π - π}{2} + x) = c o s (2 π - \frac{π}{2} + x)$

$= c o s [2 π - (\frac{π}{2} - x)]; VI quadrant.$

$c o s (\frac{π}{2} - x); i I^{st} quadrant.$

$= s i n x$

$c o s (2 x + x) = c o s x, as x lies in I^{st} quadrant.$

$c o t (2 π + x) = c o t x; as x lies in I^{st} quadrant.$

$c o t (\frac{3 π}{2} - x) = c o t [\frac{4 π - π}{2} - x]$

$= c o t [2 π - \frac{π}{2} - x]$

$= c o t [2 x - (\frac{π}{2} + x)]; V I^{th} quadrant.$

$= - c o t (\frac{π}{2} + x); I I^{nd} quadrat.$

$= - (- t a n x)$

$= t a n x .$

So.L.H.S $L.H.S = s i n x c o s x [t a n x + c o t x]$

$= s i n x c o s x [\frac{s i n x}{c o s x} + \frac{c o s x}{s i n x}]$

$= s i n x c o s x \frac{(s i n^{2} x + c o s^{2} x)}{c o s x s i n x}$

$= s i n^{2} x + c o s^{2} x$

= 1

= R.H.S.

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