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The direction cosines of normal to plane `3x-2y+6z=7` are
Solution:
Here,
`root()(a^2+b^2+c^2)=root()(3^2+2^2+6^2)=7`
Thus, the direction cosines of the normal are,
`3/7,-2/7,6/7`
The area of triangle formed by the lines `x=0`, `y+x=0`, and `x-c=0` is
If `vec a` is a non zero vector, then `(vec a)/(|vec a|)` is a
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