Advertisements
Advertisements
Question
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
Options
x − 5y + 3z = 7
x − 5y + 3z = −7
x + 5y + 3z = 7
x + 5y + 3z = −7
Solution
x − 5y + 3z = 7
\[\text{ The equation of the plane passing through the line of intersection of the given planes is } \]
\[x + y + z + 3 + \lambda \left( 2x - y + 3z + 1 \right) = 0 \]
\[\left( 1 + 2\lambda \right)x + \left( 1 - \lambda \right)y + \left( 1 + 3\lambda \right)z + 3 + \lambda = 0 . . . \left( 1 \right)\]
\[\text{ This plane is parallel to the line} \frac{x}{1}=\frac{y}{2}=\frac{z}{3} . \text{ It means that this line is perpendicular to the normal of the plane (1) } .\]
\[ \Rightarrow 1 \left( 1 + 2\lambda \right) + 2 \left( 1 - \lambda \right) + 3 \left( 1 + 3\lambda \right) = 0.................... (\text{ Because }a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]
\[ \Rightarrow 1 + 2\lambda + 2 - 2\lambda + 3 + 9\lambda = 0\]
\[ \Rightarrow 9\lambda + 6 = 0\]
\[ \Rightarrow \lambda = \frac{- 2}{3}\]
\[\text{ Substituting this in (1), we get } \]
\[\left( 1 + 2 \left( \frac{- 2}{3} \right) \right)x + \left( 1 - \left( \frac{- 2}{3} \right) \right)y + \left( 1 + 3 \left( \frac{- 2}{3} \right) \right)z + 3 + \left( \frac{- 2}{3} \right) = 0\]
\[ \Rightarrow - x + 5y - 3z + 7 = 0\]
\[ \Rightarrow x - 5y + 3z = 7\]
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 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 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 (−1, −5, −10) from the point of intersection of the line \[\vec{r} = \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) + \lambda\left( 3 \hat{i}+ 4 \hat{j} + 2 \hat{k} \right)\] and the plane \[\vec{r} . \left( \hat{i} - \hat{j} + \hat{k} \right) = 5 .\]
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 distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, −4, −5) and B(2, −3, 1) intersects the plane 2x + y + z = 7.
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 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 \[\vec{r} = 3 \hat{i} - \hat{j} - \hat{k} + \lambda\left( 2 \hat{i} - 2 \hat{j} + \hat{k} \right) .\]
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 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
The equation of the plane which cuts equal intercepts of unit length on the coordinate axes is
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.
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
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).
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.
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
