Question

# Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (− 1, 1, 2) and (− 5, − 5, − 2)

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Solution

## The vertices of a triangle is given as A( 3,5,−4 ), B( −1,1,2 ) and C( −5,−5,−2 ). The formula of direction ratios of a line passing through two points P( x,y,z ) and Q( x ′ , y ′ , z ′ ) is, ( x ′ −x ),( y ′ −y ),( z ′ −z ) Direction cosines of these points will be, ( x ′ −x ) PQ , ( y ′ −y ) PQ , ( z ′ −z ) PQ Here PQ= ( x ′ −x ) 2 + ( y ′ −y ) 2 + ( z ′ −z ) 2 The side AB has direction ratios and direction cosines as, Direction ratios=( −1−3 ),( 1−5 ),( 2−( −4 ) ) =−4,−4,6 The value of AB is, AB= ( −4 ) 2 + ( −4 ) 2 + ( 6 ) 2 = 68 =2 17 Direction cosines= −4 2 17 , −4 2 17 , 6 2 17 = −2 17 , −2 17 , 3 17 The side BC has direction ratios and direction cosines as, Direction ratios=( −5−( −1 ) ),( −5−1 ),( −2−2 ) =−4,−6,−4 The value of BC is, BC= 68 =2 17 Direction cosines= −4 2 17 , −6 2 17 , −4 2 17 = −2 17 , −3 17 , −2 17 The side CA has direction ratios and direction cosines as, Direction ratios=( 3−( −5 ) ),( −5−( 5 ) ),( −4−( −2 ) ) =8,−10,−2 The value of CA is, CA= 168 =2 42 Direction cosines= 8 2 42 , −10 2 42 , −2 2 42 = 4 42 , −5 42 , −1 42 Thus, the direction cosine of the sides of the triangle is calculated.

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