Answer:
Explanation:
(a) The distance between two points A(x1, y1, z1) and B(x2, y2, z2) is given by the formula: AB = √((x2-x1)² + (y2-y1)² + (z2-z1)²). Substituting the coordinates of points A and B into this formula gives: AB = √((2-1)² + (-1-4)² + (3-2)²) = √(14).
(b) The direction vector of line l is given by the vector (i-j+k). The direction vector of line AB is given by the vector AB = (2-1)i + (-1-4)j + (3-2)k = i - 5j + k. The cosine of the angle between two vectors is given by the dot product of the vectors divided by the product of their magnitudes. Therefore, cos(θ) = (AB.(i-j+k)) / (|AB|.|i-j+k|) = ((i - 5j + k).(i-j+k)) / (√14.√3) = -3/√42. Hence θ = cos⁻¹(-3/√42).
©(i) Let P be the point on line l where λ = p. Then P has position vector r = (2i-j+3k)+p(i-j+k) = (2+p)i + (-1-p)j + (3+p)k. The vector AP is given by AP = r - a = ((2+p)i + (-1-p)j + (3+p)k) - (i+4j+2k) = pi - (5+p)j + pk. Taking the dot product of AP with (i-j+k), we get AP.(i-j+k)=7+3p.
©(ii) The foot of the perpendicular from point A to line l is the point on line l that is closest to point A. This point is obtained when AP is perpendicular to the direction vector of line l, which is (i-j+k). Therefore, AP.(i-j+k)=0. Substituting the expression for AP.(i-j+k), we get 7+3p=0, so p=-7/3. Substituting this value of p into the expression for r, we get r = (2-7/3)i - 8/3j + 2k. Hence, the coordinates of the foot of the perpendicular from point A to line l are (1/3,-8/3,2).
(d) The cartesian equation of a line with direction vector d and passing through a point with position vector a is given by r=a+λd. For line l, d=(i-j+k), a=(2i-j+3k), so its cartesian equation is r=(2i-j+3k)+λ(i-j+k).