Find the coordinates of the foot of perpendicular drawn from the point A (1, 2, 1) to the line
fraction numerator straight x minus 1 over denominator 4 end fraction space equals space fraction numerator straight y minus 4 over denominator 0 end fraction space equals space fraction numerator straight z minus 6 over denominator negative 2 end fraction

The equation of line BC
or   fraction numerator straight x minus 1 over denominator 4 end fraction space equals space fraction numerator straight y minus 4 over denominator 0 end fraction space equals space fraction numerator straight z minus 6 over denominator negative 2 end fraction space equals space straight k

Any point D on it is
(4 k + 1, 4, – 2 k + 6)
Let D be foot of perpendicular from A on BC.
Direction ratios of AD are
4 k + 1–1, 4 – 2, – 2 k + 6 – 1 i.e., 4 k , 2, – 2 k + 5
Direction ratios of BC are 4, 0, – 2
Since AD is perpendicular to BC
∴    (4 k) (4) + (2) (0) + (– 2 k + 5) (–2) = 0
therefore space space space space space 16 straight k plus 0 plus 4 straight k minus 10 space equals space 0 space space space space space space space rightwards double arrow space space space 20 space straight k space equals space space 10 space space space space space space space rightwards double arrow space space space straight k space equals space 1 half
therefore space space space space space straight D space is space left parenthesis 2 plus 1 comma space 4 comma space minus 1 comma space plus 6 right parenthesis space straight i. straight e. comma space left parenthesis 3 comma space 4 comma space 5 right parenthesis

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Find the area of the triangle whose vertices are (1, – 1, – 3), (4, – 3, 1), (3,– 1, 2).

Let A (1, – 1, – 3), B (4, – 3, 1), C (3, – 1, 2) be vertices of Δ ABC

AB space equals space square root of left parenthesis 4 minus 1 right parenthesis squared plus left parenthesis negative 3 plus 1 right parenthesis squared plus left parenthesis 1 plus 3 right parenthesis squared end root space equals space square root of 9 plus 4 plus 16 end root space equals space square root of 29
AC space equals space square root of left parenthesis 3 minus 1 right parenthesis squared plus left parenthesis negative 1 plus 1 right parenthesis squared plus left parenthesis 2 plus 3 right parenthesis squared end root space equals space square root of 4 plus 0 plus 25 end root space equals space square root of 29

Direction-ratios of AB are 4 – 1, – 3 + 1, 1 + 3 i.e.. 3, – 2, 4 respectively.
Direction-ratios of AC are 3 – 1, –1 + 1, 2 + 3 i.e., 2, 0, 5 respectively.
therefore space space space cos space straight A space equals space fraction numerator left parenthesis 3 right parenthesis thin space left parenthesis 2 right parenthesis space plus space left parenthesis negative 2 right parenthesis space left parenthesis 0 right parenthesis space plus space left parenthesis 4 right parenthesis thin space left parenthesis 5 right parenthesis over denominator square root of 9 plus 4 plus 16 end root space square root of 4 plus 0 plus 25 end root end fraction space equals space fraction numerator 6 plus 0 plus 20 over denominator square root of 29 space square root of 29 end fraction space equals space 26 over 29
therefore space space sin space straight A space equals space square root of 1 minus space cos squared straight A end root space equals space square root of 1 minus open parentheses 26 over 29 close parentheses squared end root space equals space square root of 1 minus 676 over 841 end root space equals space square root of 165 over 841 end root space equals space fraction numerator square root of 165 over denominator 29 end fraction
Area of increment ABC space equals space 1 half AB. space AC space sin space straight A space equals space 1 half square root of 29. space square root of 29. space fraction numerator square root of 165 over denominator 29 end fraction space equals space 1 half square root of 165 space sq. space units. space

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Find the image of the point (1, 6, 3) in the line straight x over 1 space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 2 over denominator 3 end fraction.

The equations of given line are
straight x over 1 space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 2 over denominator 3 end fraction

From P (1, 6, 3) , draw PM ⊥ AB and produce it to P' (α, β γ) such that M is mid-point of PP'. Then P' is image of P in line AB.
Any point M as line AB is
(r. 2 r + 1, 3 r + 2)
Direction ratios of AB are 1, 2, 3
Direction ratios of PM are r – 1 , 2 r + 1 – 6 , 3 r + 2 – 3
i.e. r – 1, 2 r – 5, 3 r – 1
∵ PM ⊥ AB∴ (1) (r – 1) + (2) (2 r – 5) + (3) (3 r – 1) = 0
∴ i + 4 r – 10 + 9 r – 3 = 0
∴ 14 r = 14 ⇒ r = 1
∴ M is (1, 3, 5)
Now M is mid-point of PP'
therefore space space space space space fraction numerator straight alpha plus 1 over denominator 2 end fraction space equals space 1 comma space space space space fraction numerator straight beta plus 6 over denominator 2 end fraction space equals space 3 comma space space space fraction numerator straight gamma plus 3 over denominator 2 end fraction space equals space 5
therefore space space space straight alpha plus 1 space equals space 2 comma space space space space space space space space space straight beta plus 6 space equals space 6 comma space space space space straight gamma space plus space 3 space equals space 10
therefore space space space straight alpha space equals space 1 comma space space space space space space space space space space space straight beta space equals space 0 comma space space space space space space space space straight gamma space equals space 7
therefore space space space space image space is space left parenthesis 1 comma space 0 comma space 7 right parenthesis

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A line makes angle α, β, γ and δ with the diagonals of a cube, prove that
cos squared straight alpha space plus space cos squared straight beta space plus space cos squared straight gamma space space plus cos squared straight delta space equals space 4 over 3.


Take O, a corner of cube OBLCMANP, as origin and OA, OB, OC, the three edges through it as the axes.

Let OA = OB = OC = a, then the co-ordinates of O , A , B , C are (0, 0, 0), (a, 0, 0), (0, a, 0), (0, 0, a) respectively ; those of P, L, M, N are (a, a, a), (0, a, a), (a, 0, a), (a, a, 0) respectively.
The four diagonals are OP, AL, BM, CN. Direction cosines of OP are proportional to a – 0, a – 0, a – 0, i.e., a, a, a, i.e., 1,1,1.
Direction-cosines of AL are proportional to 0 – a, a – 0, a – 0 i.e., –a, a, a, i.e., – 1, 1, 1.
Direction-cosines of BM are proportional to a – 0, 0 – a, a – 0, i.e., a – a, a i.e., 1, – 1, 1.
Direction-cosines of CN are proportional to a – 0, a – 0, 0 – a i.e., a, a,– a i.e., 1, 1, – 1.
therefore  direction -cosines of OP are fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction
Directon-cosines of AL are negative fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction
Direction-cosines of BM are fraction numerator 1 over denominator square root of 3 end fraction comma space minus fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction
Direction-cosines of CN are fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction comma space minus fraction numerator 1 over denominator square root of 3 end fraction
Let l, m, n be direction-cosines of the line
∴    the line makes an angle α with OP.
therefore              cos space straight alpha space equals space straight l open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses plus straight m space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses space plus space straight n open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses
or                 cos space straight alpha space equals space fraction numerator straight l plus straight m plus straight n over denominator square root of 3 end fraction                                         ...(1)
Similarly        cos space straight beta space equals space fraction numerator negative straight l plus straight m plus straight n over denominator square root of 3 end fraction                                     ...(2)
                 cos space straight gamma space equals space fraction numerator straight l minus straight m plus straight n over denominator square root of 3 end fraction                                            ...(3)
cos space straight delta space equals space fraction numerator straight l plus straight m minus straight n over denominator square root of 3 end fraction

Squaring and adding (1), (2), (3) and (4), we get,
cos2 α + cos2 β + cos2 γ + cos2 δ
                      equals space 1 third open square brackets left parenthesis straight l plus straight m plus straight n right parenthesis squared plus left parenthesis negative straight l plus straight m plus straight n right parenthesis squared plus left parenthesis straight l minus straight m plus straight n right parenthesis squared plus left parenthesis straight l plus straight m minus straight n right parenthesis squared close square brackets
space equals space 1 third open square brackets 4 space straight l squared plus space 4 space straight m squared plus space 4 space straight n squared close square brackets space equals space 4 over 3 left parenthesis straight l squared plus space straight m squared space plus space straight n squared right parenthesis space equals space 4 over 3 left parenthesis 1 right parenthesis
therefore space space space cos squared straight alpha space plus space cos squared straight beta space plus space cos squared straight gamma space plus space cos squared straight delta space equals space 4 over 3.

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Find the area of the triangle whose vertices are (1, 2, 4), (-2, 1, 2), (2, 4, -3).


Let A (1, 2, 4), B (– 2, 1, 2), C (2, 4, – 3) be vertices of Δ ABC.

AB space equals space square root of left parenthesis negative 2 minus 1 right parenthesis squared plus left parenthesis 1 minus 2 right parenthesis squared plus left parenthesis 2 minus 4 right parenthesis squared end root space equals space square root of 9 plus 1 plus 4 end root space equals space square root of 14
AC space equals space square root of left parenthesis 2 minus 1 right parenthesis squared plus left parenthesis 4 minus 2 right parenthesis squared plus left parenthesis negative 3 minus 4 right parenthesis squared end root space space equals square root of 1 plus 4 plus 49 end root space equals space square root of 54

Direction-ratios of AB are – 2, – 1, 1 – 2, 2 – 4 i.e., – 3, – 1,– 2 respectively.
Direction-ratios of AC are 2 – 1, 4 – 2, – 3 – 4 i.e., 1, 2, – 7 respectively.
cos space straight A space equals space fraction numerator left parenthesis negative 3 right parenthesis thin space left parenthesis 1 right parenthesis space plus space left parenthesis negative 1 right parenthesis thin space left parenthesis 2 right parenthesis space plus left parenthesis negative 2 right parenthesis thin space left parenthesis negative 7 right parenthesis over denominator square root of 9 plus 1 plus 4 end root square root of 1 plus 4 plus 49 end root end fraction space equals fraction numerator negative 3 minus 2 plus 14 over denominator square root of 14 space square root of 54 end fraction space equals space fraction numerator 9 over denominator square root of 14 space square root of 54 end fraction
sin space straight A space equals space square root of 1 minus cos squared straight A end root space equals space square root of 1 minus fraction numerator 81 over denominator 14 cross times 54 end fraction end root space equals space square root of fraction numerator 766 minus 81 over denominator 14 cross times 54 end fraction end root space equals fraction numerator square root of 675 over denominator square root of 14 space square root of 54 end fraction
Area space of increment thin space ABC space equals space 1 half space AB. space AC. space sin space straight A space equals space 1 half square root of 14. space square root of 54. space fraction numerator square root of 675 over denominator square root of 14 space square root of 54 end fraction
                          equals space 1 half square root of 675 space sq. space units. space




Tips: -

Note on parallelopiped and cube
(i) A parallelopipcd is a solid bounded by three pairs of parallel plane faces.
(ii) A rectangular parallelopiped is parallelopiped whose faces are all rectangles.
(iii) A cube is a parallelopiped whose faces are all squares.

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