Question

iv) The direction ratios of normal to plane
r = (i-j)+1(i+j+k)+u(i-2j+3k)
are ....
(A) 5,2,-3
(B) 5,-2,3
(C) 5,-2, -3
(D) -5,-2,-3​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

The direction ratios of normal to plane

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

(A) 5,2,-3

(B) 5,-2,3

(C) 5,-2, -3

(D) -5,-2,-3

EVALUATION

Here the given parametric equation of the plane is

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

Now the vector form of the direction ratio of the normal to the plane is

$\displaystyle \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \\ 1 & 1& 1 \\ \\ 1 & - 2 & 3 \end{vmatrix}$

$\displaystyle \implies \vec{n} = \hat{i} (3 + 2) - \hat{j}(3 - 1) + \hat{k} ( - 2 - 1)$

$\displaystyle \implies \vec{n} = 5\hat{i} - 2 \hat{j} - 3 \hat{k}$

Hence the direction ratios of normal to plane 5 , - 2 , 3

FINAL ANSWER

Hence the correct option is (C) 5 , - 2 , - 3

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Answer 2

Answer:

The direction ratios of normal to plane

r = (i-j)+1(i+j+k)+u(i-2j+3k)

are ....