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Three Dimensional Geometry

Short Answer Type

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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Long Answer Type

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.

Given points have position vectors as

2 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.

∴ points are (2, 1, 3), (– 4, 3, – 1), (5, 0, 5).
The equations of straight lines through the points (2, 1, 3) and (– 4, 3, – 1) are

fraction numerator straight x minus 2 over denominator negative 4 minus 2 end fraction space equals space fraction numerator straight y minus 1 over denominator 3 minus 1 end fraction space equals space fraction numerator straight z minus 3 over denominator negative 1 minus 3 end fraction space space space or space space fraction numerator straight x minus 2 over denominator negative 6 end fraction space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 3 over denominator negative 4 end fraction

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

The points (5, 0, 5) will lie on it

if

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

i.e.. if 1 = 1 = 1, which is true.
∴ the points (2, 1, 3), (– 4, 3, – 1), (5, 0, 5) are collinear
∴ points with position vectors

2 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 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 5 space straight i with hat on top space plus space 5 space straight k with hat on top

are collinear.
Another Method:
Let

straight a with rightwards arrow on top space equals space 2 straight i with hat on top space plus space straight j with hat on top space plus space 3 straight k with hat on top comma space space straight b with rightwards arrow on top space equals space minus 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 straight c with rightwards arrow on top space equals space 5 space straight i with hat on top space plus space 5 space straight k with hat on top.

The equation of line through two points with positions vectors

straight a with rightwards arrow on top space and space straight b with rightwards arrow on top

straight r with rightwards arrow on top space equals space straight a with rightwards arrow on top plus straight lambda open parentheses straight b with rightwards arrow on top minus straight a with rightwards arrow on top close parentheses

or space space straight r with rightwards arrow on top space equals space open parentheses 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 close parentheses space plus space straight lambda space open parentheses negative 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 space minus space 2 space straight i with hat on top space minus space straight j with hat on top space minus space space 3 space straight k with hat on top close parentheses or space space straight r with rightwards arrow on top space equals space open parentheses 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 close parentheses space plus space straight lambda space open parentheses negative 6 space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 4 space straight k with hat on top close parentheses

Now the point

straight c with rightwards arrow on top space equals space 5 space straight i with hat on top space plus space 5 space straight k with hat on top

will lie on it
if

5 space straight i with hat on top space plus space 5 space straight k with hat on top space equals space open parentheses 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 close parentheses space plus space straight lambda space open parentheses negative 6 space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 4 space straight k with hat on top close parentheses

i.e., if

3 straight i with hat on top space minus space straight j with hat on top space plus space 2 space straight k with hat on top space equals space minus 6 space straight lambda space straight i with hat on top space plus space 2 space straight lambda space straight j with hat on top space plus space 4 space straight lambda space straight k with hat on top

i.e., if

3 equals negative 6 straight lambda comma space space minus 1 space equals space 2 straight lambda comma space space space 2 space equals space minus space 4 space straight lambda

i.e., if

straight lambda space equals space minus 1 half

therefore

for

straight lambda space equals space minus 1 half comma space space space given space points space are space collinear. space

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Short Answer Type

58.

Show that the point whose position vectors are given by $negative 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top comma space space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top space space and space space 7 straight i with hat on top space minus space straight k with hat on top$ are collinear.