Find the coordinates of the point where the line through (5,1,6)and (3,4,1) crosses the YZ plane.
It is known that the equation of the line passing through the points, (x1,y1,z1) and (x2,y2,z2), is x-\(\frac{x_1}{x_2}\)-x1=y-\(\frac{y_1}{y_2}\)-y1=z-\(\frac{z_1}{z_2}\)-z1
The line passing through the points, (5,1,6), and (3,4,1) is given by,
\(\frac{x-5}{3-5}\)=\(\frac{y-1}{4-1}\)=\(\frac{z-6}{1-6}\)
⇒\(\frac{x-5}{-2}\)=\(\frac{y-1}{3}\)=\(\frac{z-6}{-5}\)=k(say)
⇒x=5-2k,y=3k+1,z=6-5k
Any point on the line is of the form (5-2k,3k+1,6-5k).
The equation of YZ-plane is x=0
Since the line passing through YZ-plane,
5-2k=0
⇒k=\(\frac{5}{2}\)
⇒3k+1
=3×\(\frac{5}{2}\)+1
=\(\frac{17}{2}\) 6-5k
=6-5×\(\frac{5}{2}\)
=\(\frac {-13}{2}\)
Therefore, the required point is (0, 17/2, -13/2).
List - I | List - II | ||
(P) | γ equals | (1) | \(-\hat{i}-\hat{j}+\hat{k}\) |
(Q) | A possible choice for \(\hat{n}\) is | (2) | \(\sqrt{\frac{3}{2}}\) |
(R) | \(\overrightarrow{OR_1}\) equals | (3) | 1 |
(S) | A possible value of \(\overrightarrow{OR_1}.\hat{n}\) is | (4) | \(\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}\) |
(5) | \(\sqrt{\frac{2}{3}}\) |
Let \(\alpha x+\beta y+y z=1\) be the equation of a plane passing through the point\((3,-2,5)\)and perpendicular to the line joining the points \((1,2,3)\) and \((-2,3,5)\) Then the value of \(\alpha \beta y\)is equal to ____
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