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
\[\text{The direction cosines of the line passing through two points }P \left( x_1 , y_1 , z_1 \right) \text{ and} \ Q \left( x_2 , y_2 , z_2 \right) \text{are} \frac{x_2 - x_1}{PQ}, \frac{y_2 - y_1}{PQ}, \frac{z_2 - z_1}{PQ} . \]\[\text{ Here,} \]
\[PQ = \sqrt{\left( x_2 - x_1 \right)^2 + \left( y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}\]
\[P = \left( - 2, 4, - 5 \right) \]
\[Q = \left( 1, 2, 3 \right)\]
\[ \therefore PQ = \sqrt{\left[ 1 - \left( - 2 \right) \right]^2 + \left( 2 - 4 \right)^2 + \left[ 3 - \left( - 5 \right) \right]^2} = \sqrt{77}\]
\[\text{Thus, the direction cosines of the line joining two points are }\frac{1 - \left( - 2 \right)}{\sqrt{77}}, \frac{2 - 4}{\sqrt{77}}, \frac{3 - \left( - 5 \right)}{\sqrt{77}}, \text{i . e }. \frac{3}{\sqrt{77}}, \frac{- 2}{\sqrt{77}}, \frac{8}{\sqrt{77}} .\]
Page 2
\[\text{The given points are} \text{ A }\left( 2, 3, - 4 \right), B\left( 1, - 2, 3 \right) \text{and}\ C \left( 3, 8, - 11 \right) . \]
\[\text{We know that the direction ratios of the line joining the points, } \left( x_1 , y_1 , z_1 \right) \text{and}\ \left( x_2 , y_2 , z_2 \right) \text{are } \ x_2 - x_1 , y_2 - y_1 , z_2 - z_1 . \]
\[\text{The direction ratios of the line joining A and B are } 1 - 2, - 2 - 3, 3 + 4,\text{ i . e }. - 1, - 5, 7 . \]
\[\text{The direction ratios of the line joining B and C are } 3 - 1, 8 + 2, - 11 - 3, \text{i . e }. 2, 10, - 14 . \]
\[\text {It is clear that the direction ratios of BC are - 2 times that of AB, i . e . they are proportional . }\]
\[\text{Therefore, AB is parallel to BC . }\]
\[\text{Also, point B is common in both AB and BC . }\]
\[\text{Therefore, points A, B and C are collinear .}\]
Text Solution
Solution : (x_1,y_1,z_1)=A and (x_2,y_2,z_2)=B<br> dintance between A and B<br> AB=`sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1))`<br> =`sqrt(9+4+64)`<br> =`sqrt77`<br> `(l,m,n)=((x_2-x_1)/(AB),(y_2-y_1)/(AB),(z_2-z_1)/(AB))`<br> after putting the values<br> =`(3/sqrt77,2/sqrt77,8/sqrt77)`
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Example, 3 Find the direction cosines of the line passing through the two points ( 2, 4, 5) and (1, 2, 3). P ( 2, 4, 5) Q (1, 2, 3) So, x1 = 2, y1 = 4 , z1 = 5 & x2 = 1, y2 = 2 , z2 = 3 Direction ratios = (x2 x1), (y2 y1), (z2 z1) = 1 ( 2) , 2 4 , 3 ( 5) = 1 + 2, 2, 3 + 5 = 3, 2, 8 Direction cosines = 3 32 + 2 2 + 82 , 2 32 + 2 2 + 82 , 8 32 + 2 2 + 82 = 3 9 + 4 + 64 , 2 9 + 4 + 64 , 8 9 + 4 + 64 = , ,