Advertisements
Advertisements
प्रश्न
Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.
Advertisements
उत्तर
\[ \text{ The equation of any plane passing through (1, -1, 2) is } \]
\[a \left( x - 1 \right) + b \left( y + 1 \right) + c \left( z - 2 \right) = 0 . . . \left( 1 \right)\]
\[ \text{ It is given that (1) is perpendicular to the plane 2x + 3y - 2z = 5 . So, } \]
\[2a + 3b - 2c = 0 . . . \left( 2 \right)\]
\[\text{ It is given that (1) is perpendicular to the plane x + 2y - 3z = 8 . So } ,\]
\[a + 2b - 3c = 0 . . . \left( 3 \right)\]
\[\text{ Solving (1), (2) and (3), we get } \]
\[\begin{vmatrix}x - 1 & y + 1 & z - 2 \\ 2 & 3 & - 2 \\ 1 & 2 & - 3\end{vmatrix} = 0\]
\[ \Rightarrow - 5 \left( x - 1 \right) + 4 \left( y + 1 \right) + 1 \left( z - 2 \right) = 0\]
\[ \Rightarrow 5x - 4y - z = 7\]
APPEARS IN
संबंधित प्रश्न
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector.`3hati + 5hatj - 6hatk`
Find the equations of the planes that passes through three points.
(1, 1, 0), (1, 2, 1), (−2, 2, −1)
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes `vecr = (hati - hatj + 2hatk) = 5`and `vecr.(3hati + hatj + hatk) = 6`
Find the vector and Cartesian equations of a plane passing through the point (1, −1, 1) and normal to the line joining the points (1, 2, 5) and (−1, 3, 1).
find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane
Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3).
Find the vector equation of the plane passing through points A (a, 0, 0), B (0, b, 0) and C(0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from the origin, prove that \[\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} .\]
Find the vector equation of the plane passing through the points \[3 \hat{i} + 4 \hat{j} + 2 \hat{k} , 2 \hat{i} - 2 \hat{j} - \hat{k} \text{ and } 7 \hat{i} + 6 \hat{k} .\]
Determine the value of λ for which the following planes are perpendicular to each ot
2x − 4y + 3z = 5 and x + 2y + λz = 5
Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.
Find the equation of the plane passing through the points whose coordinates are (−1, 1, 1) and (1, −1, 1) and perpendicular to the plane x + 2y + 2z = 5.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 .\]
Find the vector equation of the line through the origin which is perpendicular to the plane \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = 3 .\]
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the zx - plane .
Find the coordinates of the point where the line through (3, −4, −5) and (2, −3, 1) crosses the plane 2x + y + z = 7.
Find the equation of a plane which passes through the point (3, 2, 0) and contains the line \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\] .
Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line \[\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1} .\]
Hence, or otherwise, deduce the length of the perpendicular.
Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular.
Find the coordinates of the foot of the perpendicular from the point (2, 3, 7) to the plane 3x − y − z = 7. Also, find the length of the perpendicular.
Find the direction cosines of the unit vector perpendicular to the plane \[\vec{r} \cdot \left( 6 \hat{i} - 3 \hat{j} - 2 \hat{k} \right) + 1 = 0\] passing through the origin.
Find the length and the foot of perpendicular from the point \[\left( 1, \frac{3}{2}, 2 \right)\] to the plane \[2x - 2y + 4z + 5 = 0\] .
Write the equation of the plane parallel to the YOZ- plane and passing through (−4, 1, 0).
Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).
Write the ratio in which the plane 4x + 5y − 3z = 8 divides the line segment joining the points (−2, 1, 5) and (3, 3, 2).
Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.
Find the length of the perpendicular drawn from the origin to the plane 2x − 3y + 6z + 21 = 0.
The equation of the plane parallel to the lines x − 1 = 2y − 5 = 2z and 3x = 4y − 11 = 3z − 4 and passing through the point (2, 3, 3) is
Find the value of λ for which the following lines are perpendicular to each other `("x"-5)/(5λ+2) = (2 -"y")/(5) = (1 -"z")/(-1); ("x")/(1) = ("y"+1/2)/(2λ) = ("z" -1)/(3)`
hence, find whether the lines intersect or not
Find the vector equation of the plane which contains the line of intersection of the planes `vec("r").(hat"i"+2hat"j"+3hat"k"),-4=0, vec("r").(2hat"i"+hat"j"-hat"k")+5=0`and which is perpendicular to the plane`vec("r").(5hat"i"+3hat"j"-6hat"k"),+8=0`
The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by ______.
Show that the points `(hat"i" - hat"j" + 3hat"k")` and `3(hat"i" + hat"j" + hat"k")` are equidistant from the plane `vec"r" * (5hat"i" + 2hat"j" - 7hat"k") + 9` = 0 and lies on opposite side of it.
If the foot of perpendicular drawn from the origin to a plane is (5, – 3, – 2), then the equation of plane is `vec"r".(5hat"i" - 3hat"j" - 2hat"k")` = 38.
The point at which the normal to the curve y = `"x" + 1/"x", "x" > 0` is perpendicular to the line 3x – 4y – 7 = 0 is:
A unit vector perpendicular to the plane ABC, where A, B and C are respectively the points (3, –1, 2), (1, –1, –3) and (4, –3, 1), is
