Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (–1, 1, 2) and (–5, –5, –2).
Let A(3, 5, – 4), B(–1, 1, 2), C(–5, –5, –2) be the vertices of ΔABC. Direction ratios of AB are – 1 – 3, 1 – 5, 2 + 4 i.e. – 4, – 4, 6 cosines of the line AB as i.e. Direction ratios of BC are – 5 + 1, –5 –1, –2 –2 i.e. – 4, –6, –4. Dividing each by direction ratios of the line BC as Direction ratios of CA are 3+5, 5+5, -4+2 i.e., 8, 10 -2. Dividing each by direction ratios of the line CA as Choose correct alternatives : The direction ratios of the line 3x + 1 = 6y – 2 = 1 – z are ______
The direction ratios of the line 3x + 1 = 6y – 2 = 1 – z are 2, 1, – 6 Concept: Distance Between Skew Lines and Parallel Lines Is there an error in this question or solution? Given equations of the line are:3x -1 = 6y +2 = 1 - z Rewriting the above equation, we have, `3(x-1/3)=6(y+2/6)=-(z-1)` `((x-1/3))/(1/3)=((y+1/3))/(1/6)=((z-1))/-1 ....(1)` Now consider the general equation of the line: `(x-a)/l=(y-b)/m=(z-c)/n...(2)` where, l,m and n are the direction ratios of the line and the point (a,b,c) lies on the line. we have l=1/3, m=1/6 and n=-1 Also, a=1/3, b=-1/3 and c=1 This shows that the given line passes through (1/3, -1/3, 1) Therefore, the given line passes through the point having position vector `bara=1/3hati-i/3hatj+hatk` and is parallel to the So its vector equation is `barr=(1/3hati-1/3hatj+hatk)+lambda(1/3hati+1/6hatj-hatk)` |