Advertisements
Advertisements
प्रश्न
Find the image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0.
Advertisements
उत्तर
\[\text{ Let Q be the image of the point P(1, 3, 4) in the plane 2x - y + z + 3 = 0 } . \]
\[\text{ Then PQ is normal to the plane. So, the direction ratios of PQ are proportional to 2, -1, 1.} \]
\[\text{ Since PQ passes through P (1, 3, 4) and has direction ratios proportional to 2, -1 and 1 , equation of PQ is } \]
\[\frac{x - 1}{2} = \frac{y - 3}{- 1} = \frac{z - 4}{1} = r (say)\]
\[\text{ Let the coordinates of Q be } \left( 2r + 1, - r + 3, r + 4 \right). \text{ Let R be the mid-point of PQ. Then,} \]
\[R = \left( \frac{2r + 1 + 1}{2}, \frac{- r + 3 + 3}{2}, \frac{r + 4 + 4}{2} \right) = \left( r + 1, \frac{- r + 6}{2}, \frac{r + 8}{2} \right)\]
\[\text{ Since R lies in the plane } 2x - y + z + 3 = 0, \]
\[2 \left( r + 1 \right) - \left( \frac{- r + 6}{2} \right) + \frac{r + 8}{2} + 3 = 0\]
\[ \Rightarrow 4r + 4 + r - 6 + r + 8 + 6 = 0\]
\[ \Rightarrow 6r + 12 = 0\]
\[ \Rightarrow r = - 2\]
\[\text{ Substituting this in the coordinates of Q, we get } \]
\[Q = \left( 2r + 1, - r + 3, r + 4 \right) . = \left( 2 \left( - 2 \right) + 1, 2 + 3, - 2 + 4 \right) = \left( - 3, 5, 2 \right)\]
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, −1), (6, 4, −5), (−4, −2, 3)
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.
If the points (1, 1, p) and (−3, 0, 1) be equidistant from the plane `vecr.(3hati + 4hatj - 12hatk)+ 13 = 0`, then find the value of p.
Find the vector equation of each one of following planes.
2x − y + 2z = 8
Find the vector equation of each one of following planes.
x + y = 3
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 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 equation of the plane passing through the origin and perpendicular to each of the planes x + 2y − z = 1 and 3x − 4y + z = 5.
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 that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.
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 equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.
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 image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.
Find the reflection of the point (1, 2, −1) in the plane 3x − 5y + 4z = 5.
Find the image of the point with position vector \[3 \hat{i} + \hat{j} + 2 \hat{k} \] in the plane \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + \hat{k} \right) = 4 .\] Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through \[3 \hat{i} + \hat{j} + 2 \hat{k} .\]
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 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 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.
Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.
Write the equation of a plane which is at a distance of \[5\sqrt{3}\] units from origin and the normal to which is equally inclined to coordinate axes.
Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1.
Find the co-ordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1).
Find the image of the point (1, 6, 3) in the line `x/1 = (y - 1)/2 = (z - 2)/3`.
The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by ______.
Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.
Find the length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 0.
Find the equations of the line passing through the point (3, 0, 1) and parallel to the planes x + 2y = 0 and 3y – z = 0.
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`.
