Advertisements
Advertisements
प्रश्न
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).
उत्तर
The equation of any plane through the intersection of the planes, 3x − y + 2z − 4 = 0 and x + y + z − 2 = 0, is
`3x − y + 2z − 4 + alpha`(x + y + z − 2) = 0 where `alpha in R` ....(1)
The plane passes through the point (2, 2, 1). Therefore, this point will satisfy equation (1).
This is the required equation of the plane.
APPEARS IN
संबंधित प्रश्न
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 distance of the point (1, -5, 9) from the plane
Find the equation of the plane containing the line \[\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}\] and the point (0, 7, −7) and show that the line \[\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}\] also lies in the same plane.
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.
Show that the plane whose vector equation is \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - \hat{k} \right) = 3\] contains the line whose vector equation is \[\vec{r} = \hat{i} + \hat{j} + \lambda\left( 2 \hat{i} + \hat{j} + 4 \hat{k} \right) .\]
Find the equation of the plane determined by the intersection of the lines \[\frac{x + 3}{3} = \frac{y}{- 2} = \frac{z - 7}{6} \text{ and }\frac{x + 6}{1} = \frac{y + 5}{- 3} = \frac{z - 1}{2}\]
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
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 =
A vector parallel to the line of intersection of the planes\[\vec{r} \cdot \left( 3 \hat{i} - \hat{j} + \hat{k} \right) = 1 \text{ and } \vec{r} \cdot \left( \hat{i} + 4 \hat{j} - 2 \hat{k} \right) = 2\] is
The distance of the point (−1, −5, −10) from the point of intersection of the line \[\vec{r} = 2 \hat{i}- \hat{j} + 2 \hat{k} + \lambda\left( 3 \hat{i} + 4 \hat{j}+ 12 \hat{k} \right)\] and the plane \[\vec{r} \cdot \left( \hat{i} - \hat{j} + \hat{k} \right) = 5\] is
The equation of the plane through the intersection of the planes ax + by + cz + d = 0 andlx + my + nz + p = 0 and parallel to the line y=0, z=0
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.
Find the distance of the point (–1, –5, – 10) from the point of intersection of the line `vec"r" = 2hat"i" - hat"j" + 2hat"k" + lambda(3hat"i" + 4hat"j" + 2hat"k")` and the plane `vec"r" * (hat"i" - hat"j" + hat"k")` = 5
Find the vector equation of the line which is parallel to the vector `3hat"i" - 2hat"j" + 6hat"k"` and which passes through the point (1, –2, 3).
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.
Prove that the line through A(0, – 1, – 1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(– 4, 4, 4).
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.
ABCD be a parallelogram and M be the point of intersection of the diagonals, if O is any point, then OA + OB + OC + OD is equal to
The equation of straight line through the intersection of the lines x – 2y = 1 and x + 3y = 2 and parallel to 3x + 4y = 0 is
The equation of the curve passing through the point `(0, pi/4)` whose differential equation is sin x cos y dx + cos x sin y dy = 0, is
