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
Dividing each by
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
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 ______
2, 1, 6
2, 1, – 6
2, – 1, 6
– 2, 1, 6
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.
Compare the equation (1), with the general equation (2),
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
vector `barb=1/3hati+1/6hatj-hatk`
So its vector equation is
`barr=(1/3hati-1/3hatj+hatk)+lambda(1/3hati+1/6hatj-hatk)`