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Three Dimensional Geometry

Question
CBSEENMA12033243

A and B are the points (2, – 1, 3) and (4, 2, 5). Find the projection of AB on a line which is inclined at equal acute angles with the co-ordinate axes.

Solution
Let α be the angle which the line PQ makes with all the axes.
therefore  its direction cosines are cos space straight alpha comma space space cosα comma space cosα.
therefore space space cos squared straight alpha plus cos squared straight alpha plus cos squared straight alpha space equals space 1 space space space space space space space space space left square bracket straight l squared plus straight m squared plus straight n squared space equals space 1 right square bracket
therefore space space 3 space cos squared space straight alpha space equals space 1 space space space space rightwards double arrow space space space space cos squared straight alpha space equals space 1 third space space space rightwards double arrow space space space cos space straight alpha space equals space plus-or-minus fraction numerator 1 over denominator square root of 3 end fraction
therefore    cos space straight alpha space equals space fraction numerator 1 over denominator square root of 3 end fraction                        open square brackets because space line space makes space acute space angle space straight alpha space with space each space axis close square brackets
therefore                  straight l space equals straight m space equals space straight n space space equals fraction numerator 1 over denominator square root of 3 end fraction
where l, m, n are direction cosines of line PQ.
A, B are points (2, – 1, 3), (4, 2, 5).
therefore    projection of AB on PQ = (4 - 2). fraction numerator 1 over denominator square root of 3 end fraction + (2+1).  fraction numerator 1 over denominator square root of 3 end fraction plus left parenthesis 5 minus 3 right parenthesis. space fraction numerator 1 over denominator square root of 3 end fraction
                                                   open square brackets because space space of space space left parenthesis straight x subscript 2 minus straight x subscript 1 right parenthesis straight l space plus space left parenthesis straight y subscript 2 minus straight y subscript 1 right parenthesis space straight m space plus space left parenthesis straight z subscript 2 minus straight z right parenthesis space straight n close square brackets
                             equals space fraction numerator 2 over denominator square root of 3 end fraction plus fraction numerator 3 over denominator square root of 3 end fraction plus fraction numerator 2 over denominator square root of 3 end fraction space equals space fraction numerator 2 plus 3 plus 2 over denominator square root of 3 end fraction space equals fraction numerator 7 over denominator square root of 3 end fraction space units. space

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.