If direction ratios of the normal of the plane which contains the lines (x-2/3)=(y-4/2)=(z-1/1) and (x-6/3)= (y+2/2)=(z-2/1) are (a, 1,-26), then a is equal to

Question

If direction ratios of the normal of the plane which contains the lines (x-2/3)=(y-4/2)=(z-1/1) and (x-6/3)= (y+2/2)=(z-2/1) are (a, 1,-26), then a is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The given lines are parallel <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/0988bf01330ad3dd63cf22a61be53b75-.png" /> vecn =A B ×(3 hati+2 hatj+ hatk) =(4 hati-6 hatj+ hatk) ×(3 hati+2 hatj+ hatk) =-8 hati- hatj+26 hatk ⇒ Direction ratios of normal are (8,+1,-26) ⇒ a=8

Q. If direction ratios of the normal of the plane which contains the lines $\frac{x - 2}{3} = \frac{y - 4}{2} = \frac{z - 1}{1}$ and $\frac{x - 6}{3} =$ $\frac{y + 2}{2} = \frac{z - 2}{1}$ are $(a, 1, - 26)$ , then $a$ is equal to

The given lines are parallel

$n = A B \times (3 \hat{i} + 2 \hat{j} + \hat{k})$
$= (4 \hat{i} - 6 \hat{j} + \hat{k}) \times (3 \hat{i} + 2 \hat{j} + \hat{k})$
$= - 8 \hat{i} - \hat{j} + 26 \hat{k}$
$\Rightarrow$ Direction ratios of normal are $(8, + 1, - 26)$
$\Rightarrow a = 8$

Q. If direction ratios of the normal of the plane which contains the lines 3x−2​=2y−4​=1z−1​ and 3x−6​= 2y+2​=1z−2​ are (a,1,−26), then a is equal to

The given lines are parallel n=AB×(3i^+2j^​+k^) =(4i^−6j^​+k^)×(3i^+2j^​+k^) =−8i^−j^​+26k^ ⇒ Direction ratios of normal are (8,+1,−26) ⇒a=8

Q. If direction ratios of the normal of the plane which contains the lines $\frac{x - 2}{3} = \frac{y - 4}{2} = \frac{z - 1}{1}$ and $\frac{x - 6}{3} =$ $\frac{y + 2}{2} = \frac{z - 2}{1}$ are $(a, 1, - 26)$ , then $a$ is equal to

The given lines are parallel

$n = A B \times (3 \hat{i} + 2 \hat{j} + \hat{k})$
$= (4 \hat{i} - 6 \hat{j} + \hat{k}) \times (3 \hat{i} + 2 \hat{j} + \hat{k})$
$= - 8 \hat{i} - \hat{j} + 26 \hat{k}$
$\Rightarrow$ Direction ratios of normal are $(8, + 1, - 26)$
$\Rightarrow a = 8$