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प्रश्न
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`
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उत्तर
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संबंधित प्रश्न
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.
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 Cartesian form of the equation of a plane whose vector equation is
\[\vec{r} \cdot \left( 12 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + 5 = 0\]
Find the vector equation of each one of following planes.
x + y = 3
The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane.
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.
Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.
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 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 through the origin which is perpendicular to the plane \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = 3 .\]
Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis.
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 coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the zx - 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 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 coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.
Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).
Write the value of k for which the planes x − 2y + kz = 4 and 2x + 5y − z = 9 are perpendicular.
Write the equation of the plane \[\vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c}\] in scalar product form.
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 vector equation of the plane containing the line \[\vec{r} = \left( - 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \lambda\left( 3 \hat{i} - 2 \hat{j} - \hat{k} \right)\] and the point \[\hat{i} + 2 \hat{j} + 3 \hat{k} \] is
Find the image of the point (1, 6, 3) in the line `x/1 = (y - 1)/2 = (z - 2)/3`.
The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by ______.
Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 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 length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 0.
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.
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,
The coordinates of the foot of the perpendicular drawn from the point A(1, 0, 3) to the join of the points B(4, 7, 1) and C(3, 5, 3) are
