96.
Show that the points A (6. – 7, 0) B (16, – 19, – 4). C (0, 3, – 6), D (2, – 5, 10) are such that AB and CD intersect at the point P (1, – 1, 2).
We are given the points A (6, – 7, 0), B (16, – 19, – 4), C (0, 3, – 6),  D (2, – 5. 10), P (1, – 1, 2).
 ![AB with rightwards arrow on top space equals space PV space of space straight B space minus space straight P. straight V. space of space straight A space equals space left parenthesis 16 space straight i with hat on top space minus space 19 space straight j with hat on top space minus space 4 space straight k with hat on top right parenthesis space minus space left parenthesis 6 space straight i with hat on top space minus space 7 space straight j with hat on top space plus space 0 space straight k with hat on top right parenthesis
space space space space space space space equals space 10 space straight i with hat on top space minus space 12 space straight j with hat on top space minus space 4 space straight k with hat on top
AP with rightwards arrow on top space equals space straight P. straight V. space of space straight P space minus space straight P. straight V. space of space straight A space equals space left parenthesis straight i with hat on top space minus space straight j with hat on top space minus space space 2 space straight k with hat on top right parenthesis space minus space left parenthesis 6 space straight i with hat on top space minus space 7 space straight j with hat on top space plus space 0 space straight k with hat on top right parenthesis
space space space space space space space equals space minus space 5 space straight i with hat on top space plus space 6 space straight j with hat on top space space plus space 2 space straight k with hat on top
PB with rightwards arrow on top space equals space straight P. straight V. space of space straight B space minus space straight P. straight V. space of space straight P space equals space left parenthesis 16 space straight i with hat on top space minus space 19 space straight j with hat on top space minus space 4 space straight k with hat on top right parenthesis space minus space left parenthesis 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 right parenthesis
space space space space space space space equals space 5 space straight i with hat on top space minus space 18 space straight j with hat on top space minus space 6 space straight k with hat on top
therefore space space AB space equals space open vertical bar AB with rightwards arrow on top close vertical bar space equals space square root of 100 plus 144 plus 16 end root space equals space square root of 260 space equals space square root of 4 space cross times space 65 end root space equals space 2 square root of 65
space space space space space space AP space equals space open vertical bar AP with rightwards arrow on top close vertical bar space equals space square root of 25 plus 36 plus 4 end root space equals square root of 65
space space space space space space PB space equals space open vertical bar PB with rightwards arrow on top close vertical bar space equals space square root of 225 plus 324 plus 36 end root space equals space square root of 585 space equals square root of 9 space cross times space 65 end root space equals 3 square root of 65](/application/zrc/images/qvar/MAEN12066655.png)
![therefore space space space space BA plus space AP space equals space BP
therefore space space space space straight P comma space straight A comma space straight B space are space collinear.](/application/zrc/images/qvar/MAEN12066655-1.png)
Again ![CD with rightwards arrow on top space equals space straight P. straight V. space of space straight D space minus space straight P. straight V. space of space straight C space equals space left parenthesis 2 straight i with hat on top space minus space 5 space straight j with hat on top space plus space 10 space straight k with hat on top right parenthesis space minus space left parenthesis 3 space straight j with hat on top space minus space 6 space straight k with hat on top right parenthesis](/application/zrc/images/qvar/MAEN12066655-2.png)
          ![equals space 2 space straight i with hat on top space minus space 8 space straight j with hat on top space plus space 16 space straight k with hat on top](/application/zrc/images/qvar/MAEN12066655-3.png)
       ![CP with rightwards arrow on top space equals space straight P. straight V. space of space straight P space minus space straight P. straight V. space of space straight C space equals space left parenthesis 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 right parenthesis space minus space left parenthesis 3 space straight j with hat on top space minus space 6 space straight k with hat on top right parenthesis
space space space space space space space equals space straight i with hat on top space minus space 4 space straight j with hat on top space plus space 8 space straight k with hat on top](/application/zrc/images/qvar/MAEN12066655-4.png)
      ![PD with rightwards arrow on top space equals space straight P. straight V. space of space straight D space minus space straight P. straight V. space of space straight P space equals space open parentheses 2 space straight i with hat on top space minus space 5 space straight j with hat on top space plus space 10 space straight k with hat on top close parentheses space minus space left parenthesis 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 right parenthesis
space space space space space space equals space straight i with hat on top space minus space 4 space straight j with hat on top space plus space 8 space straight k with hat on top](/application/zrc/images/qvar/MAEN12066655-5.png)
![therefore space space space space CD space equals open vertical bar CD with rightwards arrow on top close vertical bar space equals space square root of 4 plus 64 plus 256 end root space equals square root of 324 space equals 18
space space space space space space space space CP space equals space open vertical bar CP with rightwards arrow on top close vertical bar space equals space square root of 1 plus 16 plus 64 end root space equals space square root of 81 equals space 9
space space space space space space space space space PD equals space open vertical bar PD with rightwards arrow on top close vertical bar space equals space square root of 1 plus 16 plus 64 end root space equals square root of 81 space equals 9
therefore space space space space CP plus PD space equals space CD space space space space rightwards double arrow space space space space straight P comma space straight C comma space straight D space are space collinear.
therefore space space space space AB space and space CD space intersect space at space straight P left parenthesis 1 comma space minus 1 comma space space 2 right parenthesis.](/application/zrc/images/qvar/MAEN12066655-6.png)
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