Question

Find a unit vector in the direction ū if ū is the vector from P(2,1, -3) to ((-1,0,4). Then, find c such that vector PR is orthogonal to ū where Ręc, c,c).

Answer 1

Given P(2,1, -3) and Q(-1,0, 4) 7 is a vector from P to Q => 7 = position vector of Q - position vector of P => 7 = ((-1)i +0j + 4k) - (21+ (1)j. +(-3)k) => 7=(-1 - 2)i + (0 - 1)j + (4+3)k => 7 = -3i - j + 7k ū unit vector in direction of 7 => unit vector in direction of -3i - j + 7k V-3)2 + (-1)2 + (7) -31 - 1 + 7k (9+1+49) => unit vector in direction of 7 => unit vector in direction of 7 -3i - j + 7k ✓(59) 1 => unit vector in direction of 7 = vo vai+ 7 k /59 Answer.
$To\ find\ c\ such\ that\ \overrightarrow{PR}\ is\ orthogonal \ to\ \overrightarrow{v} \\ \\ firstly\ we\ will\ find\ \overrightarrow{PR} \\ \\ \overrightarrow{PR} = Position\ vector\ of\ R-position\ vector\ of\ P \\ \\ \\ =>\overrightarrow{PR} =(ci+cj+ck) -(2i +j-3k)\\ \\ \\=>\overrightarrow{PR} =((c-2)i +(c-1)j+(c+3)k) \\ \\ \\ Since\ \overrightarrow{PR} \ is\ perpendicular\ to\overrightarrow{v} \\ \\ \\ => \overrightarrow{PR} \cdot \overrightarrow{v}=0 \\ \\ \\ =>((c-2)i +(c-1)j+(c+3)k) \cdot (-3i-j+7k) =0 \\ \\ \\=>(c-2)(-3)+(c-1)(-1)+(c+3)(7)=0 \\ \\ \\ =>-3c+6-c+1+7c+21=0 \\ \\ =>3c+28=0 \\ \\ c= \frac{-28}{3} \\ \\ Answer$