Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
\[\text{ The equation of any plane passing through (1, -3, -2) is } \]
\[a \left( x - 1 \right) + b \left( y + 3 \right) + c \left( z + 2 \right) = 0 . . . \left( 1 \right)\]
\[\text{ It is given that (1) is perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8 . Then } ,\]
\[a + 2b + 2c = 0 . . . \left( 2 \right)\]
\[3a + 3b + 2c = 0 . . . \left( 3 \right)\]
\[\text{ Solving (1), (2) and (3), we get } \]
\[\begin{vmatrix}x - 1 & y + 3 & z + 2 \\ 1 & 2 & 2 \\ 3 & 3 & 2\end{vmatrix} = 0\]
\[ \Rightarrow - 2 \left( x - 1 \right) + 4 \left( y + 3 \right) - 3 \left( z + 2 \right) = 0\]
\[ \Rightarrow - 2x + 2 + 4y + 12 - 3z - 6 = 0\]
\[ \Rightarrow 2x - 4y + 3z - 8 = 0\]
APPEARS IN
संबंधित प्रश्न
Find the equations of the planes that passes through three points.
(1, 1, 0), (1, 2, 1), (−2, 2, −1)
Find the equation of the plane passing through (a, b, c) and parallel to the plane `vecr.(hati + hatj + hatk) = 2`
If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the plane passing through P and perpendicular to OP.
Find the vector equation of a plane passing through a point with position vector \[2 \hat{i} - \hat{j} + \hat{k} \] and perpendicular to the vector \[4 \hat{i} + 2 \hat{j} - 3 \hat{k} .\]
Find the vector equations of the coordinate planes.
Find the vector equation of each one of following planes.
2x − y + 2z = 8
The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane.
A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point
Show that the normals to the following pairs of planes are perpendicular to each other.
x − y + z − 2 = 0 and 3x + 2y − z + 4 = 0
Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.
Find the vector equation of a plane which is at a distance of 3 units from the origin and has \[\hat{k}\] as the unit vector normal to it.
Find the vector equation of the plane passing through the points (1, 1, 1), (1, −1, 1) and (−7, −3, −5).
Find the vector equation of the plane passing through the points \[3 \hat{i} + 4 \hat{j} + 2 \hat{k} , 2 \hat{i} - 2 \hat{j} - \hat{k} \text{ and } 7 \hat{i} + 6 \hat{k} .\]
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 the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
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 equation of the plane passing through the intersection of the planes x − 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane
If the lines \[\frac{x - 1}{- 3} = \frac{y - 2}{- 2k} = \frac{z - 3}{2} \text{ and }\frac{x - 1}{k} = \frac{y - 2}{1} = \frac{z - 3}{5}\] are perpendicular, find the value of k and, hence, find the equation of the plane containing these lines.
Find the reflection of the point (1, 2, −1) in the plane 3x − 5y + 4z = 5.
Find the distance of the point (1, −2, 3) from the plane x − y + z = 5 measured along a line parallel to \[\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6} .\]
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 coordinates of the foot of the perpendicular and the perpendicular distance of the point P (3, 2, 1) from the plane 2x − y + z + 1 = 0. Also, find the image of the point in the plane.
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 coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.
Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence, find the distance of point P (–2, 5, 5) from the plane obtained
Write the distance of the plane \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) = 12\] from the origin.
Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y −z = 7.
The vector equation of the plane containing the line \[\vec{r} = \left( - 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \lambda\left( 3 \hat{i} - 2 \hat{j} - \hat{k} \right)\] and the point \[\hat{i} + 2 \hat{j} + 3 \hat{k} \] 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 image of the point (1, 6, 3) in the line `x/1 = (y - 1)/2 = (z - 2)/3`.
Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′, respectively, from the origin, prove that `1/"a"^2 + 1/"b"^2 + 1/"c"^2 = 1/"a'"^2 + 1/"b'"^2 + 1/"c'"^2`
The equation of a line, which is parallel to `2hat"i" + hat"j" + 3hat"k"` and which passes through the point (5, –2, 4), is `(x - 5)/2 = (y + 2)/(-1) = (z - 4)/3`.
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:
Let A be the foot of the perpendicular from focus P of hyperbola `x^2/a^2 - y^2/b^2 = 1` on the line bx – ay = 0 and let C be the centre of hyperbola. Then the area of the rectangle whose sides are equal to that of PA and CA is,
