Advertisements
Advertisements
प्रश्न
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane
Advertisements
उत्तर
It is known that the equation of the line passing through the points, (x1, y1, z1) and (x2, y2, z2), is
The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,
Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k).
The equation of YZ-plane is x = 0
Since the line passes through YZ-plane,
5 − 2k = 0
APPEARS IN
संबंधित प्रश्न
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
2x + 3y – z = 5
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0
Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX 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).
The planes: 2x − y + 4z = 5 and 5x − 2.5y + 10z = 6 are
(A) Perpendicular
(B) Parallel
(C) intersect y-axis
(C) passes through `(0,0,5/4)`
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.
Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal 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.
Prove that the line of section of the planes 5x + 2y − 4z + 2 = 0 and 2x + 8y + 2z − 1 = 0 is parallel to the plane 4x − 2y − 5z − 2 = 0.
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.
Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line \[\vec{r} = \hat{i} + 3 \hat{j} - 2 \hat{k} + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) .\]
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 \[\vec{r} = \hat{i} - \hat{j} + \lambda\left( \hat{i} + \hat{j} + \hat{k} \right) + \mu\left( \hat{i} - 2 \hat{j} + 3 \hat{k} \right)\] in scalar product form is
Find the image of the point having position vector `hat"i" + 3hat"j" + 4hat"k"` in the plane `hat"r" * (2hat"i" - hat"j" + hat"k") + 3` = 0.
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.
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
