Find the distance of a point (2, 5, – 3)...

Find the distance of a point (2, 5, – 3) from the plane `vecr cdot(6hati-3hatj+2hatk) = 4`

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Find the perpendicular distance from the point (1,2,3) to the line : vecr=6hati+7hatj+7hatk+lambda(3hati+2hatj-2hatk)

Find the image of the point hati+3hatj+4hatk in the plane vecr.(2hati-hatj+hatk)+3=0 .

Find the direction cosines of the unit vector perpendicular to the plane vecr cdot(6hati-3hatj-2hatk) +1 = 0 passing through the origin.

Find the equation of the plane which contains the line of intersection of the planes vecr cdot(hati+2hatj+3hatk) - 4 = 0 and vecr cdot (2hati + hatj - hatk) +5 = 0 and which is perpendicular to the plane vecr cdot (5hati+3hatj-6hatk) +8 =0

Find the equation of the plane passing through the line of intersection of the planes vecr cdot (hati+hatj+hatk) = 1 and vecr cdot(2hati+3hatj-hatk)+4 = 0 and parallel to x-axis.

Find the distance of the point (2, 3, -1) from the point of intersection of the line vecr=3hati+hatj-2hatk+lambda(3hati+6hatj+2hatk) and the plane vecr.(2hati-hatj+hatk) =3.

Find the distance of the point (-1 , -5 , -10) from the point of intersection of the line vecr=2hati-hatj+2hatk+lambda(3hati+4hatj+2hatk) and the plane vecr.(hati-hatj+hatk) =5.

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 equation of the plane passing through (a, b, c) and parallel to the plane vecr cdot(hati + hatj + hatk) = 2

Find the distance of the point (2,12,5) from the point of intersection of the line: vecr=2hati-4hatj+2hatk+lambda(3hati+4hatj+2hatk) and the plane vecr*(hati-2hatj+hatk)=0 .

PSEB-THREE DIMENSIONAL GEOMETRY-Exercise

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