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51. Find a vector equation of the line through the points A (3, 4, – 7) and B (1, – 1, 6).

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52. Find the vector equation for the line passing through the points (– 1, 0, 2) and (3, 4, 6).

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53. Find the vector and cartesian equations of the line that passes through the origin and (5,- 2, 3).

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54. Find the vector and the cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).

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55. Find the vector equation of line joining the points whose vectors are

2 space straight i with hat on top minus straight j with hat on top plus straight k with hat on top space space and space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top.

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56. If the points (–1, 3, 2), (– 4, 2, –2) and (5, 5, λ) are collinear, then find the value of λ.

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57. Show that the points whose position vectors are given by

2 space straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top comma space space minus space 4 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space straight k with hat on top comma space space 5 space straight i with hat on top space plus space 5 space straight k with hat on top

are collinear.

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Given points have position vectors as

negative 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top comma space space space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top space space and space 7 space straight i with hat on top space minus space straight k with hat on top.

∴ points are (–2, 3, 5), (1, 2, 3), (7, 0, –1)
The equation of st. line through (–2, 3, 5), (1, 2, 3) is

fraction numerator straight x plus 2 over denominator 1 plus 2 end fraction space equals space fraction numerator straight y minus 3 over denominator 2 minus 3 end fraction space equals space fraction numerator straight z minus 5 over denominator 3 minus 5 end fraction space space space space space or space space space fraction numerator straight x plus 2 over denominator 3 end fraction space equals space fraction numerator straight y minus 3 over denominator negative 1 end fraction space equals space fraction numerator straight z minus 5 over denominator negative 2 end fraction

The point (7, 0, –1) will lie on it
if

fraction numerator 7 plus 2 over denominator 3 end fraction space equals space fraction numerator 0 minus 3 over denominator negative 1 end fraction space equals space fraction numerator negative 1 minus 5 over denominator negative 2 end fraction

i.e. if 3 = 3 = 3, which is true.
∴ the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.
∴ points with position vectors

negative 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top comma space space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top

and

7 space straight i with hat on top space minus space straight k with hat on top

are collinear.

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59. Find the coordinates of the point where the line through A(3, 4, 1) and B (5, 1, 6) crosses the x y-plane.

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60. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.

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