Tag: vectors:planes in three dimensions
Questions Related to vectors:planes in three dimensions
The cartesian equation of the plane which is at a distance of 10 unite from the original and perpendicular to the vector i + 2j -2k is
The equation of the plane through the point $(0, -1, -6)$ and $(-2, 9, 3)$ are perpendicular to the plane $x-4y-2z=8$ is
The normal form of $2x-2y+z=5$ is
If a line is given by $\dfrac{x-2}3 = \dfrac{y+10}5 = \dfrac{z+6}2$, then which of the following points lies on this line?
Vector form of plane $2x-z+1=0$ is _________
Find the equation of the plane through the points $(1, 0, -1), (3, 2, 2)$ and parallel to the line $\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-2}{3}$.
The equation of the plane passing through the straight line $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-3}{4}$ and perpendicular to plane $x+2y +z=12$ is:
Equation of the plane containing the straight lines $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}$ and perpendicular to the plane containing the straight lines $\dfrac{x}{3} = \dfrac{y}{4} = \dfrac{z}{2}$ and $\dfrac{x}{4} = \dfrac{y}{2} = \dfrac{z}{3}$
Let a,b,c be any real numbers.Suppose that there are real numbers x,y,z not all zero such that $x=cy+bz , y=az+cx$ and $z=bx+ay$, then ${a^2} + {b^2} + {c^2} + 2abc $ is equal to
The direction cosines of the normal to the plane $x+2y-3z+4=0$ are