Question

iv) The direction ratios of normal to plane
r = (i-j)+2 (i+j+k)+u(i-2j+3k)
are
(A) 5,2,-3
(C) 5,-2,-3
(B) 5,-2,3
(D) –5,-2,-3
2
2

Answer 1

Step-by-step explanation:

Given plane

$\vec{r} = ( \hat{i} - \hat{j}) + 2(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k})$

The given vector is parallel to $\vec{a} =\hat{i}+\hat{j}+\hat{k}\:\: \& \: \: \vec{b}=\hat{i} - 2\hat{j}+3\hat{k}$

so, the given vector will be perpendicular to $\vec{a}\times\vec{b}$

$\vec{a} \times \vec{b} = 5 \hat{i} - 2\hat{j} -3 \hat{k}$

So required equation of plane is

$\{\vec{r} - (\hat{i} - \hat{j}) \}.(5\hat{i} - 2\hat{j} - 3\hat{k}) = 0$

Dr of normal is (5, -2, -3)