Advertisements
Advertisements
Question
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.
Advertisements
Solution
\[\text { Let M be the foot of the perpendicular of the origin P(0, 0, 0) in the plane 2x - 3y + 4z - 6 = 0 .} \]
\[\text{ Then,PM is normal to the plane. So, the direction ratios of PM are proportional to 2, -3, 4. } \]
\[\text{ Since PM passes through P (0, 0, 0) and has direction ratios proportional to 2, -3 , 4 , the equation of PQ is } \]
\[\frac{x - 0}{2} = \frac{y - 0}{- 3} = \frac{z - 0}{4} = r (\text{ say } )\]
\[\text{ Let the coordiantes of M be } \left( 2r, - 3r, 4r \right).\]
\[\text{ Since M lies in the plane 2x - 3y + 4z - 6 = 0, } \]
\[2 \left( 2r \right) - 3 \left( - 3r \right) + 4 \left( 4r \right) - 6 = 0\]
\[ \Rightarrow 4r + 9r + 16r - 6 = 0\]
\[ \Rightarrow 29r - 6 = 0\]
\[ \Rightarrow r = \frac{6}{29}\]
\[\text{ Substituting the value of r in the coordinates of M, we get } \]
\[M = \left( 2r, - 3r, 4r \right) = \left( 2 \left( \frac{6}{29} \right), - 3 \left( \frac{6}{29} \right), 4 \left( \frac{6}{29} \right) \right) = \left( \frac{12}{29}, \frac{- 18}{29}, \frac{24}{29} \right)\]
APPEARS IN
RELATED QUESTIONS
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 Cartesian form of the equation of a plane whose vector equation is
\[\vec{r} \cdot \left( 12 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + 5 = 0\]
Find the vector equation of each one of following planes.
2x − y + 2z = 8
Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.
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
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 P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3).
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
Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.
Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.
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 passing through the point (1, −1, 2) and perpendicular to the plane 2x − y + 3z − 5 = 0.
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 vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - 5 \hat{k} \right) + 9 = 0 .\]
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the zx - plane .
Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane \[\vec{r} \cdot \left( \hat{i} - 2 \hat{j} + 4 \hat{k} \right) + 5 = 0 .\]
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 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
Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector \[2 \hat{i} + 3 \hat{j} + 4 \hat{k} \] to the plane \[\vec{r} . \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) - 26 = 0\] Also find image of P in the plane.
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.
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 vector equation of the plane, passing through the point (a, b, c) and parallel to the plane \[\vec{r} . \left( \hat{i} + \hat{j} + \hat{k} \right) = 2\]
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 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`
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.
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,
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
