Find the vector equation of the plane passing through three points with (i + j - 2k), (2i - j + k))

Question

Find the vector equation of the plane passing through three points with \(\hat{i}+\hat{j}-2\hat{k},2\hat{i}-\hat{j}+\hat{k}\) and \(\hat{i}+2\hat{j}+\hat{k}.\) Also find the coordinates of the point of intersection of this plane and the line \(\vec{r}=3\hat{i}-\hat{j}-\hat{k}+ \lambda (2\hat{i}-2\hat{j}+\hat{k}).\)

Answer 1

The equation of plane passing through three points \(\hat{i}+\hat{j}-2\hat{k},2\hat{i}-\hat{j}+\hat{k}\) and \(\hat{i}+2\hat{j}+\hat{k}\)

Its vector form is \(\vec{r} . (9\hat{i} + 3\hat{j} - \hat{k}) = 14\)

The given line is \(\vec{r}=(3\hat{i}-\hat{j}-\hat{k})+ \lambda (2\hat{i}-2\hat{j}+\hat{k}).\)

Its cartesian form is

Let the line (ii) intersect plane (i) at \((\alpha, \beta, \gamma)\)

\(\because\) \((\alpha, \beta, \gamma)\) lie on (ii)

Also, point \((\alpha, \beta, \gamma)\) lie on plane (i)

Therefore, point of intersection \(\equiv\) (1, 1, - 2).