Advertisements
Advertisements
Question
Find the equation of the plane passing through the intersection of the planes `vecr . (hati + hatj + hatk)` and `vecr.(2hati + 3hatj - hatk) + 4 = 0` and parallel to the x-axis. Hence, find the distance of the plane from the x-axis.
Solution
We have,
`vecr.(hati + hatj + hatk) - 1 = 0` ....(1)
`vecr.(2hati + 3hatj - hatk) + 4 = 0` .....(2)
Equation of plane passing through the intersection of the planes (i) and (ii), is given by
`[vecr.(hati + hatj + hatk) - 1 ] = λ[vecr. (2hati + 3hatj - hatk) + 4 ] = 0`
⇒ `vecr.[(1 + 2λ)hati + (1 + 3λ)hatj + (1 - λ)hatk] - 1 + 4λ = 0` ....(iii)
If plane (iii) is parallel to x-axis, then
1 + 2λ = 0 ⇒ λ = `-1/2`
Therefore, the equation of the required plane is `vecr.(-hatj + 3hatk) = 6`
Distance of the plane `vecr.(-hatj + 3hatk) = 6` from x-axis is given by,
`d = ((1 xx 0 - 3 xx 0 + 6)/(sqrt(1^2 + 3^2))) = 6/sqrt10`
APPEARS IN
RELATED QUESTIONS
Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).
Find the vector equation of the plane passing through the intersection of the planes `vecr.(2hati + 2hatj - 3hatk) = 7, vecr.(2hati + 5hatj + 3hatk) = 9` and through the point (2, 1, 3)
Find the equation of the plane passing through the line of intersection of the planes `vecr.(hati + hatj + hatk) = 1` and `vecr.(2hati + 3hatj -hatk) + 4 = 0` and parallel to x-axis.
Find the equation of the plane which contains the line of intersection of the planes `vecrr.(hati + 2hatj + 3hatk) - 4 = 0, vecr.(2hati + htj - hatk) + 5 = 0`, and which is perpendicular to the plane `vecr.(5hati + 3hatj - 6hatk) + 8 = 0`.
Find the equation of the plane through the intersection of the planes 3x − 4y + 5z = 10 and 2x + 2y − 3z = 4 and parallel to the line x = 2y = 3z.
Find the coordinates of the point where the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}\] intersects the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane.
Find the distance of the point (2, 12, 5) from the point of intersection of the line \[\vec{r} = 2 \hat{i} - 4 \hat{j}+ 2 \hat{k} + \lambda\left( 3 \hat{i} + 4 \hat{j} + 2 \hat{k} \right)\] and \[\vec{r} . \left( \hat{i} - 2 \hat{j} + \hat{k} \right) = 0\]
Find the distance of the point P(−1, −5, −10) from the point of intersection of the line joining the points A(2, −1, 2) and B(5, 3, 4) with the plane \[x - y + z = 5\] .
Find the equation of the plane which contains two parallel lines\[\frac{x - 4}{1} = \frac{y - 3}{- 4} = \frac{z - 2}{5}\text{ and }\frac{x - 3}{1} = \frac{y + 2}{- 4} = \frac{z}{5} .\]
Show that the lines \[\frac{x + 4}{3} = \frac{y + 6}{5} = \frac{z - 1}{- 2}\] and 3x − 2y + z + 5 = 0 = 2x + 3y + 4z − 4 intersect. Find the equation of the plane in which they lie and also their point of intersection.
Find the coordinates of the point where the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}\] intersect the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane.
Find the vector equation of the plane passing through three points with position vectors \[\hat{i} + \hat{j} - 2 \hat{k} , 2 \hat{i} - \hat{j} + \hat{k} \text{ and } \hat{i} + 2 \hat{j} + \hat{k} .\] Also, find the coordinates of the point of intersection of this plane and the line \[\vec{r} = 3 \hat{i} - \hat{j} - \hat{k} + \lambda\left( 2 \hat{i} - 2 \hat{j} + \hat{k} \right) .\]
Find the distance of the point with position vector
The plane 2x − (1 + λ) y + 3λz = 0 passes through the intersection of the planes
The equation of the plane through the intersection of the planes x + 2y + 3z = 4 and 2x + y − z = −5 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is
The equation of the plane through the line x + y + z + 3 = 0 = 2x − y + 3z + 1 and parallel to the line \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] is
A plane meets the coordinate axes at A, B and C such that the centroid of ∆ABC is the point (a, b, c). If the equation of the plane is \[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = k,\] then k =
Show that the lines `("x"-1)/(3) = ("y"-1)/(-1) = ("z"+1)/(0) = λ and ("x"-4)/(2) = ("y")/(0) = ("z"+1)/(3)` intersect. Find their point of intersection.
Show that the lines `(x - 1)/2 = (y - 2)/3 = (z - 3)/4` and `(x - 4)/5 = (y - 1)/2` = z intersect. Also, find their point of intersection.
Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is `"a"x + "b"y +- (sqrt("a"^2 + "b"^2) tan alpha)`z = 0.
Find the equation of the plane through the intersection of the planes `vec"r" * (hat"i" + 3hat"j") - 6` = 0 and `vec"r" * (3hat"i" - hat"j" - 4hat"k")` = 0, whose perpendicular distance from origin is unity.
Find the equation of line parallel to the y-axis and drawn through the point of intersection of x – 4y + 1 = 0 and 2x + y – 7 = 0.
