Advertisements
Advertisements
प्रश्न
Find the value of λ such that the line \[\frac{x - 2}{6} = \frac{y - 1}{\lambda} = \frac{z + 5}{- 4}\] is perpendicular to the plane 3x − y − 2z = 7.
उत्तर
\[\text{ Direction ratios of the given line are proportional to 6, } \lambda, -4.\]
\[\text{ Direction ratios of the plane are 3, -1, -2.}\]
\[\text{ Since the given line is parallel to the given plane, the line is perpendicular to the normal of the given plane. } \]
\[ \Rightarrow \left( 6 \right) \left( 3 \right) + \left( \lambda \right) \left( - 1 \right) + \left( - 4 \right) \left( - 2 \right) = 0\]
\[ \Rightarrow 18 - \lambda + 8 = 0\]
\[ \Rightarrow \lambda = 26\]
APPEARS IN
संबंधित प्रश्न
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
z = 2
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
x + y + z = 1
Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX − plane.
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 coordinates of the point where the line through the points (3, - 4, - 5) and (2, - 3, 1), crosses the plane determined by the points (1, 2, 3), (4, 2,- 3) and (0, 4, 3)
Find the equation of the plane passing through the point (2, 3, 1), given that the direction ratios of the normal to the plane are proportional to 5, 3, 2.
If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP.
Reduce the equation \[\vec{r} \cdot \left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) + 6 = 0\] to normal form and, hence, find the length of the perpendicular from the origin to the plane.
Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.
The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane.
Find a unit normal vector to the plane x + 2y + 3z − 6 = 0.
Find the equation of a plane which is at a distance of \[3\sqrt{3}\] units from the origin and the normal to which is equally inclined to the coordinate axes.
Find the vector equation of the plane which is at a distance of \[\frac{6}{\sqrt{29}}\] from the origin and its normal vector from the origin is \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} .\] Also, find its Cartesian form.
Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.
Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line \[\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{- 1}\]
Write the plane \[\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) = 14\] in normal form.
Write a vector normal to the plane \[\vec{r} = l \vec{b} + m \vec{c} .\]
Write the vector equation of the line passing through the point (1, −2, −3) and normal to the plane \[\vec{r} \cdot \left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right) = 5 .\]
The equation of the plane containing the two lines
The equation of the plane \[\vec{r} = \hat{i} - \hat{j} + \lambda\left( \hat{i} + \hat{j} + \hat{k} \right) + \mu\left( \hat{i} - 2 \hat{j} + 3 \hat{k} \right)\] in scalar product form is
The equations of x-axis in space are ______.
Find the equation of a plane which is at a distance `3sqrt(3)` units from origin and the normal to which is equally inclined to coordinate axis.
If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane.
The plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis. The value of α is equal to ______.
The unit vector normal to the plane x + 2y +3z – 6 = 0 is `1/sqrt(14)hat"i" + 2/sqrt(14)hat"j" + 3/sqrt(14)hat"k"`.
Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector `3hati + 5hatj - 6hatk`
What will be the cartesian equation of the following plane. `vecr * (hati + hatj - hatk)` = 2
In the following cases find the c9ordinates of foot of perpendicular from the origin `2x + 3y + 4z - 12` = 0
Find the vector and cartesian equations of the planes that passes through (1, 0, – 2) and the normal to the plane is `hati + hatj - hatk`
