Question

The sine of the angle between the straight line (x-2)/(3)-(y-3)/(4)=(z-4)/(5) and the plane 2x-2y+1-5 is​

Answer 1

Appropriate Question :-

The sine of the angle between the straight line

$\rm \: \dfrac{x - 2}{3} = \dfrac{y - 3}{4} = \dfrac{z - 4}{5} \: and \: plane \: 2x - 2y + z - 5 = 0$

is ______

$\large\underline{\sf{Solution-}}$

Given equation of line is

$\rm \: \dfrac{x - 2}{3} = \dfrac{y - 3}{4} = \dfrac{z - 4}{5}$

So, direction ratios of the line be (3, 4, 5).

So, in vector form, direction ratio of the line is

$\rm \: \vec{b} = 3\hat{i} + 4\hat{j} + 5\hat{k}$

Also, given equation of plane is

$\rm \: 2x - 2y + z - 5 = 0$

So, normal to the surface of plane is (2, - 2, 1).

So, normal vector to the plane is

$\rm \: \vec{n} = 2\hat{i} - 2\hat{j} + \hat{k}$

We know, Angle between the line and plane is given by

$\boxed{ \rm{ \:sin \theta \: = \: \dfrac{\vec{b} \: . \: \vec{n}}{ |\vec{b}| \: |\vec{n}| } \: }}$

So,

$\rm \: \vec{b}.\vec{n}$

$\rm \: = \: (3\hat{i} + 4\hat{j} + 5\hat{k}).(2\hat{i} - 2\hat{j} + \hat{k})$

$\rm \: = \: 6 - 8 + 5$

$\rm \: = \: 3$

Now,

$\rm \: |\vec{b}|$

$\rm \: = \: |3\hat{i} + 4\hat{j} + 5\hat{k}|$

$\rm \: = \: \sqrt{ {3}^{2} + {4}^{2} + {5}^{2} }$

$\rm \: = \: \sqrt{9 + 16 + 25}$

$\rm \: = \: \sqrt{50}$

$\rm \: = \: 5 \sqrt{2}$

Now,

$\rm \: |\vec{n}|$

$\rm \: = \: |2\hat{i} - 2\hat{j} + \hat{k}|$

$\rm \: = \: \sqrt{ {(2)}^{2} + {( - 2)}^{2} + {1}^{2} }$

$\rm \: = \: \sqrt{4 + 4 + 1}$

$\rm \: = \: \sqrt{9}$

$\rm \: = \: 3$

So, on substituting the values, we get

$\rm \: sin \theta \: = \: \dfrac{3}{5 \sqrt{2} \times 3}$

$\red{\rm\implies \:\boxed{ \rm{ \:\rm \: sin \theta \: = \: \dfrac{1}{5 \sqrt{2}} \: }}}$

Answer 2

Step-by-step explanation:

Appropriate Question :-

The sine of the angle between the straight line

\begin{gathered}\rm \: \dfrac{x - 2}{3} = \dfrac{y - 3}{4} = \dfrac{z - 4}{5} \: and \: plane \: 2x - 2y + z - 5 = 0 \\ \end{gathered}

3

x−2

=

4

y−3

=

5

z−4

andplane2x−2y+z−5=0

is ______

\large\underline{\sf{Solution-}}

Solution−

Given equation of line is

\begin{gathered}\rm \: \dfrac{x - 2}{3} = \dfrac{y - 3}{4} = \dfrac{z - 4}{5} \\ \end{gathered}

3

x−2

=

4

y−3

=

5

z−4

So, direction ratios of the line be (3, 4, 5).

So, in vector form, direction ratio of the line is

\begin{gathered}\rm \: \vec{b} = 3\hat{i} + 4\hat{j} + 5\hat{k} \\ \end{gathered}

b

=3

i

^

+4

j

^

+5

k

^

Also, given equation of plane is

\begin{gathered}\rm \: 2x - 2y + z - 5 = 0 \\ \end{gathered}

2x−2y+z−5=0

So, normal to the surface of plane is (2, - 2, 1).

So, normal vector to the plane is

\begin{gathered}\rm \: \vec{n} = 2\hat{i} - 2\hat{j} + \hat{k} \\ \end{gathered}

n

=2

i

^

−2

j

^

+

k

^

We know, Angle between the line and plane is given by

\begin{gathered}\boxed{ \rm{ \:sin \theta \: = \: \dfrac{\vec{b} \: . \: \vec{n}}{ |\vec{b}| \: |\vec{n}| } \: }} \\ \end{gathered}

sinθ=

∣

b

∣∣

n

∣

b

.

n

So,

\begin{gathered}\rm \: \vec{b}.\vec{n} \\ \end{gathered}

b

.

n

\begin{gathered}\rm \: = \: (3\hat{i} + 4\hat{j} + 5\hat{k}).(2\hat{i} - 2\hat{j} + \hat{k}) \\ \end{gathered}

=(3

i

^

+4

j

^

+5

k

^

).(2

i

^

−2

j

^

+

k

^

)

\begin{gathered}\rm \: = \: 6 - 8 + 5 \\ \end{gathered}

=6−8+5

\begin{gathered}\rm \: = \: 3 \\ \end{gathered}

=3

Now,

\begin{gathered}\rm \: |\vec{b}| \\ \end{gathered}

∣

b

∣

\begin{gathered}\rm \: = \: |3\hat{i} + 4\hat{j} + 5\hat{k}| \\ \end{gathered}

=∣3

i

^

+4

j

^

+5

k

^

∣

\begin{gathered}\rm \: = \: \sqrt{ {3}^{2} + {4}^{2} + {5}^{2} } \\ \end{gathered}

=

3

2

+4

2

+5

2

\begin{gathered}\rm \: = \: \sqrt{9 + 16 + 25} \\ \end{gathered}

=

9+16+25

\begin{gathered}\rm \: = \: \sqrt{50} \\ \end{gathered}

=

50

\begin{gathered}\rm \: = \: 5 \sqrt{2} \\ \end{gathered}

=5

2

Now,

\begin{gathered}\rm \: |\vec{n}| \\ \end{gathered}

∣

n

∣

\begin{gathered}\rm \: = \: |2\hat{i} - 2\hat{j} + \hat{k}| \\ \end{gathered}

=∣2

i

^

−2

j

^

+

k

^

∣

\rm \: = \: \sqrt{ {(2)}^{2} + {( - 2)}^{2} + {1}^{2} } =

(2)

2

+(−2)

2

+1

2

\begin{gathered}\rm \: = \: \sqrt{4 + 4 + 1} \\ \end{gathered}

=

4+4+1

\begin{gathered}\rm \: = \: \sqrt{9} \\ \end{gathered}

=

9

\begin{gathered}\rm \: = \: 3 \\ \end{gathered}

=3

So, on substituting the values, we get

\begin{gathered}\rm \: sin \theta \: = \: \dfrac{3}{5 \sqrt{2} \times 3} \\ \end{gathered}

sinθ=

5

2

×3

3

\begin{gathered} \red{\rm\implies \:\boxed{ \rm{ \:\rm \: sin \theta \: = \: \dfrac{1}{5 \sqrt{2}} \: }}} \\ \end{gathered}

⟹

sinθ=

5

2

1

thanks