Advertisements
Advertisements
प्रश्न
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.
पर्याय
True
False
उत्तर
This statement is True.
Explanation:
The given equation of the plane is `vec"r".(5hat"i" - 3hat"j" - 2hat"k")` = 38
If the foot of the perpendicular to this plane is (5, – 3, – 2)
i.e., `5hat"i" - 3hat"j" - 2hat"k"` then
`(5hat"i" - 3hat"j" - 2hat"k").(5hat"i" - 3hat"j" - 2hat"k")` = 38
⇒ 25 + 9 + 4 = 38
38 = 38 .....(satisfied)
APPEARS IN
संबंधित प्रश्न
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes `vecr = (hati - hatj + 2hatk) = 5`and `vecr.(3hati + hatj + hatk) = 6`
Find the vector equation of a plane passing through a point with position vector \[2 \hat{i} - \hat{j} + \hat{k} \] and perpendicular to the vector \[4 \hat{i} + 2 \hat{j} - 3 \hat{k} .\]
Find the vector equation of each one of following planes.
x + y − z = 5
Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.
Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector \[\hat{i} - \text{2 } \hat{j} - \text{2 } \hat{k} .\]
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 \[3 \hat{i} + 4 \hat{j} + 2 \hat{k} , 2 \hat{i} - 2 \hat{j} - \hat{k} \text{ and } 7 \hat{i} + 6 \hat{k} .\]
Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.
Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y − z = 1 and 3x − 4y + z = 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 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 (a, b, c) and parallel to the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 .\]
Find the equation of a plane which passes through the point (3, 2, 0) and contains the line \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\] .
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.
Write the equation of the plane parallel to the YOZ- plane and passing through (−4, 1, 0).
Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).
Write the equation of the plane \[\vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c}\] in scalar product form.
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.
The equation of the plane parallel to the lines x − 1 = 2y − 5 = 2z and 3x = 4y − 11 = 3z − 4 and passing through the point (2, 3, 3) is
Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
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 image of the point (1, 6, 3) in the line `x/1 = (y - 1)/2 = (z - 2)/3`.
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.
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`
Find the foot of perpendicular from the point (2, 3, –8) to the line `(4 - x)/2 = y/6 = (1 - z)/3`. Also, find the perpendicular distance from the given point to the line.
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.
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 method of splitting a single force into two perpendicular components along x-axis and y-axis is called as ______.