Given - \(\bar{r}\) = (2\(\hat{i}\) + 5\(\hat{j}\) + 7\(\hat{k}\)) + λ(\(\hat{i}\) + 3\(\hat{j}\) + 4\(\hat{k}\)) and \(\bar{r}\). (\(\hat{i}\) + \(\hat{j}\) - \(\hat{k}\)) = 7.
To prove – The line and the plane are parallel &
To find – The distance between them
Direction ratios of the line = (1, 3, 4)
Direction ratios of the normal of the plane = (1, 1, - 1)
Formula to be used – If (a, b, c) be the direction ratios of a line and (a’, b’, c’) be the direction ratios of the normal to the plane, then, the angle between the two is given by sin-1\((\cfrac{axa'+bxb'+cxc'}{\sqrt{a^2+b^2+c^2}\sqrt{a^{'^{2}}+b^{'^2}+c^{'^2}}})\)
The angle between the line and the plane
Hence, the line and the plane are parallel.
Now, the equation of the plane may be written as x + y - z = 7.
Tip – If ax + by + c + d = 0 be a plane and \(\bar{r}\) = (a'\(\hat{i}\) + b'\(\hat{j}\) + c'\(\hat{k}\)) + λ(a"\(\hat{i}\) + b"\(\hat{j}\) + c"\(\hat{k}\)) be a line vector, then the distance between them is given by \(|\cfrac{axa'+bxb'+cxc'+d}{\sqrt{a^2+b^2+c^2}}|\)
The distance between the plane and the line
