given that vector a = 3i+2j+2k and vector b=1i-2j-4k are the position vectors of points P and Q respectively - Maths - Vector Algebra

Question

given that vector a = 3i+2j+2k and vector b=1i-2j-4k are the
position vectors of points P and Q respectively Find the equation
for the plane passing thorugh Q and perpendicular to the line PQ
What is the distance from point (1,-1,1) to the plane

Manbar Singh · Accepted Answer

Dear Student,

We are given that position vectors of point P and point Q are :a→ = 3i^ + 2j^ + 2k^b→ = i^ - 2j^ - 4k^Now, coordinates of P and Q are P3, 2, 2 and Q1, -2, -4
Now, direction ratios of line PQ are : 3-1, 2+2, 2+4≡2, 4, 6We know that equation of a plane passing through point x1, y1, z1 is ax-x1 + by-y1 + cz-z1 = 0where a,b,c ≡ direction ratios of the normal to the plane
Now, the required plane is ⊥ to line AB, so we havedr's of normal to required plane = dr's of line ABSo, a,b,c = 2, 4, 6Now, equation of plane passing through Q1, -2, -4 is    2x-1 + 4y+2 + 6z+4 = 0⇒2x - 2 +4y + 8 + 6z + 24 = 0⇒2x + 4y + 6z + 30 = 0⇒x + 2y + 3z + 15 = 0Now, distance of the point 1, -1, 1 from the plane x + 2y + 3z + 15 = 0 is1×1 + 2×-1 + 3×1+1512+22+32=1-2+3+1514=1714 units

Regards

given that vector a = 3i+2j+2k and vector b=1i-2j-4k are the position vectors of points P and Q respectively. Find the equation for the plane passing thorugh Q and perpendicular to the line PQ. What is the distance from point (1,-1,1) to the plane.