Advertisements
Advertisements
Question
Write the position vector of the point where the line \[\vec{r} = \vec{a} + \lambda \vec{b}\] meets the plane \[\vec{r} . \vec{n} = 0 .\]
Advertisements
Solution
\[\text{ Given equation of the line is } \]
\[ \vec{r} = a^\to + \lambda \vec{b} . . . \left( 1 \right)\]
\[\text{ Given equation of the plane is } \]
\[ \vec{r} . \vec{n} = 0\]
\[ \Rightarrow \left( \vec{a} + \lambda \vec{b} \right) . \vec{n} = 0................. [\text{ From } (1)]\]
\[ \Rightarrow \vec{a} . \vec{n} + \lambda \vec{b} . \vec{n} = 0\]
\[ \Rightarrow \lambda = - \left( \frac{\vec{a} . \vec{n}}{\vec{b} . \vec{n}} \right)\]
\[\text{ Substituting this in (1), we get} \]
\[ \vec{r} = \vec{a} - \left( \frac{\vec{a} . \vec{n}}{\vec{b} . \vec{n}} \right) \vec{b} , \text{ which is the required position vector that lies both on the line and the plane }.\]
APPEARS IN
RELATED QUESTIONS
Find the equations of the planes that passes through three points.
(1, 1, 0), (1, 2, 1), (−2, 2, −1)
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 the line passing through the point (1, 2, − 4) and perpendicular to the two lines:
`(x -8)/3 = (y+19)/(-16) = (z - 10)/7 and (x - 15)/3 = (y - 29)/8 = (z- 5)/(-5)`
Find the vector equations of the coordinate planes.
\[\vec{n}\] is a vector of magnitude \[\sqrt{3}\] and is equally inclined to an acute angle with the coordinate axes. Find the vector and Cartesian forms of the equation of a plane which passes through (2, 1, −1) and is normal to \[\vec{n}\] .
A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point
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 other.
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 the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9.
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 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 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 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 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 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.
Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).
Write the intercepts made by the plane 2x − 3y + 4z = 12 on the coordinate axes.
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).
Find the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane \[\vec{r} . \left( \hat{i} + \hat{j} + \hat{k} \right) = 2\]
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.
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 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`
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 equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
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.
`vec"AB" = 3hat"i" - hat"j" + hat"k"` and `vec"CD" = -3hat"i" + 2hat"j" + 4hat"k"` are two vectors. The position vectors of the points A and C are `6hat"i" + 7hat"j" + 4hat"k"` and `-9hat"j" + 2hat"k"`, respectively. Find the position vector of a point P on the line AB and a point Q on the line Cd such that `vec"PQ"` is perpendicular to `vec"AB"` and `vec"CD"` both.
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`.
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:
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,
