Advertisements
Advertisements
Question
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.
Solution
\[\text{ We know that the lines } \]
\[\frac{x - x_1}{l_1} = \frac{y - y_1}{m_1} = \frac{z - z_1}{n_1} \text{ and }\frac{x - x_2}{l_2} = \frac{y - y_2}{m_2} = \frac{z - z_2}{n_2} \text{ are perpendicular if } \]
\[ l_1 l_2 + m_1 m_2 + n_1 n_2 = 0\]
\[\text{ Here } ,\]
\[ l_1 = - 3; m_1 = - 2k; n_1 = 2; l_2 = k; m_2 = 1; n_2 = 5\]
\[\text{ It is given that given lines are perpendicular } .\]
\[ \Rightarrow l_1 l_2 + m_1 m_2 + n_1 n_2 = 0\]
\[ \Rightarrow \left( - 3 \right) \left( k \right) + \left( - 2k \right) \left( 1 \right) + \left( 2 \right) \left( 5 \right) = 0\]
\[ \Rightarrow - 3k - 2k + 10 = 0\]
\[ \Rightarrow - 5k = - 10\]
\[ \Rightarrow k = 2\]
\[\text{ Substituting this value in the given equations of the lines, we get } \]
\[\frac{x - 1}{- 3} = \frac{y - 2}{- 4} = \frac{z - 3}{2} . . . \left( 1 \right)\]
\[ \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 3}{5} . . . \left( 2 \right)\]
\[\text{ Finding the equation of the plane } \]
\[\text{ Let the direction ratios of the required plane be proportional to a, b, c . } \]
\[\text{ We know from (1) and (2) that lines (1) and (2) pass through the point (1, 2, 3) and the direction ratios of (1) and (2) are proportional to -3, -4, 2 and 2, 1, 5 respectively.} \]
\[\text{ Since the plane contains the lines (1) and (2), the plane must pass through the point (1, 2, 3) and it must be parallel to the line. } \]
\[\text{ So, the equation of the plane is } \]
\[a \left( x - 1 \right) + b \left( y - 2 \right) + c \left( z - 3 \right) = 0 . . . \left( 3 \right)\]
\[ - 3a - 4b + 2c = 0 . . . \left( 4 \right)\]
\[2a + b + 5c = 0 . . . \left( 5 \right)\]
\[\text{ Solving (1), (2) and (3), we get} \]
\[\begin{vmatrix}x - 1 & y - 2 & z - 3 \\ - 3 & - 4 & 2 \\ 2 & 1 & 5\end{vmatrix} = 0\]
\[ \Rightarrow - 22 \left( x - 1 \right) + 19 \left( y - 2 \right) + 5 \left( z - 3 \right) = 0\]
\[ \Rightarrow - 22x + 19y + 5z = 31\]
APPEARS IN
RELATED QUESTIONS
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 vector equations of the coordinate planes.
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 − z = 5
A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point
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.
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 (1, 1, −1), (6, 4, −5) and (−4, −2, 3).
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 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 image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.
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 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 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 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 passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).
Write the general equation of a plane parallel to X-axis.
Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y −z = 7.
Write the equation of the plane containing the lines \[\vec{r} = \vec{a} + \lambda \vec{b} \text{ and } \vec{r} = \vec{a} + \mu \vec{c} .\]
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 a vector of magnitude 26 units normal to the plane 12x − 3y + 4z = 1.
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 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 length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 0.
`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.
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.
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,
The method of splitting a single force into two perpendicular components along x-axis and y-axis is called as ______.