Let a line l pass through the origin and be perpendicular to the lines l1: r1 = i + j + 7k + (i + 2j + 3k), R l2: r2 = -i + j + 2k + (i + 2j + k), R If P is the point of intersection of l1 and l2,…

Given: The lines are l1: r = (1,1,7) + (1,2,3) and l2: r = (-1,1,2) + (1,2,1) A line l passes through the origin and is perpendicular to both these lines. Point P is the intersection of l1 and l2, and Q(a,b,) is the foot of the perpendicular from P on l. Find: The value of (a+b+). From the solution, the intersection point is obtained first and then the foot of the perpendicular is found on the required line. The extracted working concludes that a+b+ = 5 Therefore, the correct option is A.

Let a line l pass through the origin and be perpendicular to the… — Solution | JEE Mathematics

MCQMediumJEE 2023Equation of Line in 3D

Let a line $l$ pass through the origin and be perpendicular to the lines $l_1: \vec{r}_1 = i + j + 7k + \lambda(i + 2j + 3k), \quad \lambda \in \mathbb{R}$ $l_2: \vec{r}_2 = -i + j + 2k + \mu(i + 2j + k), \quad \mu \in \mathbb{R}$ If $P$ is the point of intersection of $l_1$ and $l_2$ , and $Q (a, b, \gamma)$ is the foot of perpendicular from $P$ on $l$ , then $(a + b + \gamma)$ is equal to:

A
$5$
B
$7$
C
$6$
D
$9$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The lines are

l_1: \vec{r} = (1,1,7) + \lambda(1,2,3)

and

l_2: \vec{r} = (-1,1,2) + \mu(1,2,1)

A line $l$ passes through the origin and is perpendicular to both these lines. Point $P$ is the intersection of $l_1$ and $l_2$ , and $Q(a,b,\gamma)$ is the foot of the perpendicular from $P$ on $l$ .

Find: The value of $(a+b+\gamma)$ .

From the solution, the intersection point is obtained first and then the foot of the perpendicular is found on the required line. The extracted working concludes that

a+b+\gamma = 5

Therefore, the correct option is A.

Common mistakes

Assuming that the line $l$ has the same direction ratios as one of the given lines is incorrect, because $l$ must be perpendicular to both given direction vectors. Instead, its direction should be obtained from a vector perpendicular to both lines.
Making an error while finding the intersection point of $l_1$ and $l_2$ leads to a wrong point $P$ . Solve the coordinate equations carefully by equating the parametric forms of both lines.
Using the coordinates of $P$ directly as the foot $Q$ is wrong, because $Q$ lies on line $l$ and is the projection of $P$ on $l$ . Use the projection relation to locate the foot correctly.

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Answer

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