If the equation of the line passing through the point (0, -1/2, 0) and perpendicular to the lines r = λ (i + aj + bk) and r = (i - j - 6k) + μ (-bi + aj + 5k) is (x-1)/(-2) = (y+4)/d = (z-c)/(-4), then a+b+c+d is equal to :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q16 – JEE-MAIN 2025

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Home Maths 3D GeometryQuestion #28

If the equation of the line passing through the point

(0, - \frac{1}{2}, 0)

and perpendicular to the lines

\vec{r} = λ (\hat{i} + a \hat{j} + b \hat{k})

and

\vec{r} = (\hat{i} - \hat{j} - 6 \hat{k}) + μ (- b \hat{i} + a \hat{j} + 5 \hat{k})

\frac{x - 1}{- 2} = \frac{y + 4}{d} = \frac{z - c}{- 4}

, then

a + b + c + d

is equal to :

Clue 1 of 3: Direction Vectors and Perpendicularity

The direction vector of a line perpendicular to two other lines is parallel to the cross product of their direction vectors, or equivalently, its dot product with each of the other line's direction vectors is zero.

🧠 Full Solution Path

Step 1: Identify Direction Vectors and Perpendicularity Conditions✦ Active

The direction vector of the required line is $\vec{v} = - 2 \hat{i} + d \hat{j} - 4 \hat{k}$ . The direction vectors of the two lines it is perpendicular to are $\vec{v_{1}} = \hat{i} + a \hat{j} + b \hat{k}$ and $\vec{v_{2}} = - b \hat{i} + a \hat{j} + 5 \hat{k}$ . Since $\vec{v}$ is perpendicular to $\vec{v_{1}}$ and $\vec{v_{2}}$ , their dot products must be zero:

\vec{v} \cdot \vec{v_{1}} = (- 2) (1) + (d) (a) + (- 4) (b) = 0 ⟹ - 2 + a d - 4 b = 0 (1)

\vec{v} \cdot \vec{v_{2}} = (- 2) (- b) + (d) (a) + (- 4) (5) = 0 ⟹ 2 b + a d - 20 = 0 (2)

Step 2: Solve for

a

b

, and

d

using the dot product conditions○ Expand

Subtract Equation (1) from Equation (2) to eliminate $a d$ :

(2 b + a d - 20) - (- 2 + a d - 4 b) = 0

2 b + a d - 20 + 2 - a d + 4 b = 0

6 b - 18 = 0 ⟹ 6 b = 18 ⟹ b = 3

Substitute $b = 3$ into Equation (1):

- 2 + a d - 4 (3) = 0

- 2 + a d - 12 = 0 ⟹ a d - 14 = 0 ⟹ a d = 14

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation when subtracting equations to avoid sign errors.

Step 3: Use the given point to find

c

and

d

, then

a

○ Expand

The line passes through the point $(0, - 1 / 2, 0)$ . Substitute these coordinates into the line equation $\frac{x - 1}{- 2} = \frac{y + 4}{d} = \frac{z - c}{- 4}$ :

\frac{0 - 1}{- 2} = \frac{- 1 / 2 + 4}{d} = \frac{0 - c}{- 4}

\frac{1}{2} = \frac{7 / 2}{d} = \frac{c}{4}

From $\frac{1}{2} = \frac{7 / 2}{d}$ , we get $d = 2 \times \frac{7}{2} = 7$ . From $\frac{1}{2} = \frac{c}{4}$ , we get $c = 2$ . Now, using $a d = 14$ and $d = 7$ , we find $a (7) = 14 ⟹ a = 2$ . Finally, calculate the sum:

a + b + c + d = 2 + 3 + 2 + 7 = 14

💡 Teacher's Secret Hint

Remember that any point on the line must satisfy its equation. This is a crucial step to find the individual values of $c$ and $d$ .

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Maths Question 16 – JEE-MAIN 2025

Clue 1 of 3: Direction Vectors and Perpendicularity

🧠 Full Solution Path