CBSE > Class 12 > Mathematics | CBSE Board | Homework Help

Ask questions which are clear, concise and easy to understand.

Ask Question

two finite set having m and n elements the total number of subset of the first set is 56 more than the total number of subsets of the second set find the value of m and n

Posted by Manshi Rana 6 years, 2 months ago

CBSE > Class 12 > Mathematics

1 answers

Rajat Barwar 6 years, 2 months ago

m=6 and n=3. you can check that only two exponents of 2 has a difference of 56.

1Thank You

ANSWER

Find the equation of a plane through the points(1,2,1) and perpendicular to the line joining the points(1,4,2)and (2,3,5).Also find the distance of the plane from the line x+3/2=y-5/-1=z-7/-1

Posted by Anjali Dubey 6 years, 2 months ago

CBSE > Class 12 > Mathematics

2 answers

Sia ? 6 years, 2 months ago

we know that ,
equation of plane passing through {tex}(x_1 ,y_1, z_1){/tex} having Direction ratios a, b,c is a {tex}(x - x_1) + b (y- y_1) + c (z - z_1) = 0{/tex}
Equation of plane passing through the point A (1, 2, 1) is given as
{tex}a (x - 1 ) + b (y- 2) + c (z - 1 ) = 0{/tex} .....(i)
Now, DR's of line PQ where {tex}P(1, 4, 2)\ and\ Q(2, 3, 5)\ are\ ( 2 -1, 3 - 4, 5-2)=(1, -1, 3).{/tex}
Plane (i) is perpendicular to line PQ.
{tex}\therefore{/tex} Direction ratios of plane (i) are (1, -1, 3) [{tex}\because{/tex} Direction ratios normal to the plane are proportional]
i.e. {tex}a = 1, b = -1, c = 3{/tex}
On putting values of a, b and c in Eq. (i), we get the required equation of plane as
{tex}1 (x- 1 ) -1 (y - 2) + 3 (z- 1 ) = 0{/tex}
{tex}x - 1 - y + 2 + 3 z - 3 = 0{/tex}
{tex}x - y + 3z - 2 = 0{/tex} ......(ii)
Now, the given equation of line is
{tex}\frac { x + 3 } { 2 } = \frac { y - 5 } { - 1 } = \frac { z - 7 } { - 1 }{/tex}....(iii)
Direction ratios of this line are {tex}(2, -1, -1){/tex} and passing through the point {tex}(-3, 5, 7){/tex}.
Direction ratios of normal to the plane (ii) are {tex}(1,-1, 3){/tex}.
To check whether the line is perpendicular to the plane , we use the condition {tex}a_1 a_2 + b_1 b_2 + c_1 c_2 = 0{/tex}
{tex}2 (1 ) -1 (-1) -1 (3) = 2 + 1 -3 = 0{/tex}
So, line (iii) is perpendicular to plane (i).
Distance of the point {tex}(-3, 5, 7){/tex} from the plane (ii)
{tex}\Rightarrow d = \left| \frac { ( - 3 ) ( 1 ) + ( 5 ) ( - 1 ) + 7 ( 3 ) - 2 } { \sqrt { ( 1 ) ^ { 2 } + ( - 1 ) ^ { 2 } + ( 3 ) ^ { 2 } } } \right|{/tex}{tex}\left[ \begin{array} { c } { \because \text { distance of the point } \left( x _ { 1 } , y _ { 1 } , z _ { 1 } \right) } \ { \text { to the plane } a x + b y + c z + d = 0 \text { is } } \\ { d = \frac { \left| a x _ { 1 } + b y _ { 1 } + a _ { 1 } + d \right| } { \sqrt { a ^ { 2 } + b ^ { 2 } + c ^ { 2 } } } } \end{array} \right]{/tex}
{tex}= \left| \frac { - 3 - 5 + 21 - 2 } { \sqrt { 1 + 1 + 9 } } \right|{/tex}
{tex}= \left| \frac { 11 } { \sqrt { 11 } } \right| = \left| \frac { ( \sqrt { 11 } ) ^ { 2 } } { \sqrt { 11 } } \right|{/tex}
{tex}= \sqrt { 11 }units{/tex}

1Thank You

Anjali Dubey 6 years, 2 months ago

Tq bhai

myCBSEguide

myCBSEguide

Test Generator

CUET Mock Tests

Download myCBSEguide App

All Courses

Other Websites

CBSE Courses

Resources

Corporate

NCERT Solutions

CBSE Sample Papers