JEE Main 2025 April

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Animated Solution for Mathematics - Three Dimensional Geometry: If the equation of the line passing through the point $(0, - 1/2, 0)$ and perpendicular to the lines $r = λ (\hat{i} + a \hat{j} + b \hat{k})$ and $r = (\hat{i} - \hat{j} - 6 \hat{k}) + μ (- b \hat{i} + a \hat{j} + 5 \hat{k})$ is $\frac{x - 1}{- 2} = \frac{y + 4}{d} = \frac{z - c}{- 4}$ , then $a + b + c + d$ is equal to :

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Progressive Hints

Hint 1

Hint 1: Visualize the 3D space. You have a target line

L

that is perpendicular to two other lines,

L_{1}

and

L_{2}

. This means the direction vector of

L

must be orthogonal to the direction vectors of both

L_{1}

and

L_{2}

. Think about what this implies for the dot product.

Hint 2

Hint 2: Recall the fundamental property of perpendicular vectors: if two vectors are perpendicular, their dot product is zero. Use the direction vectors

v_{1} = ⟨ 1, a, b ⟩

v_{2} = ⟨ - b, a, 5 ⟩

, and

v_{L} = ⟨ - 2, d, - 4 ⟩

to set up two independent linear equations.

Hint 3

Hint 3: Don't forget the point

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

. Since this point lies on the line

L

, it must satisfy the symmetric equation

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

. Substitute these coordinates to find the values of

c

and

d

directly.

Visualized Solution

Visualizing the 3D Geometry

Target line $L$ passes through $P (0, - \frac{1}{2}, 0)$ .
$L$ is perpendicular to two given lines, $L_{1}$ and $L_{2}$ .

Extracting Direction Ratios

Direction of $L_{1}$ : $v_{1} = ⟨ 1, a, b ⟩$
Direction of $L_{2}$ : $v_{2} = ⟨ - b, a, 5 ⟩$
Direction of $L$ : $v_{L} = ⟨ - 2, d, - 4 ⟩$

Dot Product Condition for $L ⊥ L_{1}$

Since $L ⊥ L_{1}$ , their direction vectors are orthogonal.
$v_{L} \cdot v_{1} = 0$
$(- 2) (1) + (d) (a) + (- 4) (b) = 0$

First Linear Equation

$- 2 + a d - 4 b = 0$
$a d - 4 b = 2$ (Equation 1)

Dot Product Condition for $L ⊥ L_{2}$

Similarly, $L ⊥ L_{2} ⟹ v_{L} \cdot v_{2} = 0$
$(- 2) (- b) + (d) (a) + (- 4) (5) = 0$

Second Linear Equation

$2 b + a d - 20 = 0$
$a d + 2 b = 20$ (Equation 2)

Eliminating $a d$ to find $b$

Equation 1: $a d - 4 b = 2$
Equation 2: $a d + 2 b = 20$
Subtracting Eq 1 from Eq 2:
$(a d + 2 b) - (a d - 4 b) = 20 - 2$
$6 b = 18 ⟹ b = 3$

Point Satisfies the Line Equation

The point $P (0, - \frac{1}{2}, 0)$ lies on the target line $L$ .
Equation of $L$ : $\frac{x - 1}{- 2} = \frac{y + 4}{d} = \frac{z - c}{- 4}$
Substitute $x = 0, y = - \frac{1}{2}, z = 0$ :
$\frac{0 - 1}{- 2} = \frac{- \frac{1}{2} + 4}{d} = \frac{0 - c}{- 4}$

Equating Parts to Find $d$

$\frac{- 1}{- 2} = \frac{\frac{7}{2}}{d} = \frac{- c}{- 4}$
$\frac{1}{2} = \frac{7}{2 d}$
$2 d = 14 ⟹ d = 7$

Equating Parts to Find $c$

Using the first and third parts of the equality:
$\frac{1}{2} = \frac{- c}{- 4}$
$\frac{1}{2} = \frac{c}{4}$
$2 c = 4 ⟹ c = 2$

Back-Substitution for $a$

We know $b = 3$ and $d = 7$ .
Substitute into Equation 1: $a d - 4 b = 2$
$a (7) - 4 (3) = 2$
$7 a - 12 = 2 ⟹ 7 a = 14 ⟹ a = 2$

Calculating $a + b + c + d$

We have found all variables:
$a = 2, b = 3, c = 2, d = 7$
Sum $= 2 + 3 + 2 + 7 = 14$

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The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

To define the line

L

, we require a point and a direction vector. We are given the anchor point

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

The line

L

is perpendicular to two lines,

L_{1}

and

L_{2}

. Their direction vectors are:

v_{1} = ⟨ 1, a, b ⟩

v_{2} = ⟨ - b, a, 5 ⟩

The target line

L

is given by the symmetric form:

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

Thus, the direction vector of

L

v_{L} = ⟨ - 2, d, - 4 ⟩

The Master Equations

Since

L ⊥ L_{1}

, the dot product

v_{L} \cdot v_{1} = 0

(- 2) (1) + (d) (a) + (- 4) (b) = 0 \Rightarrow a d - 4 b = 2

Since

L ⊥ L_{2}

, the dot product

v_{L} \cdot v_{2} = 0

(- 2) (- b) + (d) (a) + (- 4) (5) = 0 \Rightarrow a d + 2 b = 20

Subtracting the first equation from the second eliminates

a d

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

6 b = 18 \Rightarrow b = 3

Solving for Parameters

The point

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

lies on line

L

. Substituting these coordinates into the symmetric equation:

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

This simplifies to:

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

Equating the first and second parts:

\frac{1}{2} = \frac{7}{2 d} \Rightarrow 2 d = 14 \Rightarrow d = 7

Equating the first and third parts:

\frac{1}{2} = \frac{c}{4} \Rightarrow 2 c = 4 \Rightarrow c = 2

Final Calculation

Substitute

d = 7

and

b = 3

into the first equation

a d - 4 b = 2

a (7) - 4 (3) = 2

7 a - 12 = 2 \Rightarrow 7 a = 14 \Rightarrow a = 2

The values are

a = 2, b = 3, c = 2, d = 7

. The sum of these values is:

2 + 3 + 2 + 7 = 14

Common Traps & FAQs

Wait, why can't we just use the cross product of the direction vectors to find the direction of the target line?

You absolutely can! The direction vector of the target line is parallel to

v_{1} \times v_{2}

. However, in this specific problem, the direction vector of the target line is already partially given as

⟨ - 2, d, - 4 ⟩

. Using the dot product method is often faster here because it directly creates the linear equations needed to solve for the unknown variables

a

b

, and

d

without needing to compute a full cross product.

Why do we substitute the point

P

into the line equation? Does it really have to satisfy it?

Yes, absolutely! By definition, a line is the locus of all points

(x, y, z)

that satisfy its equation. If a line passes through a point

P

, then the coordinates of

P

must make the symmetric equation true. It is the 'anchor' that fixes the line in space. Without using this point, you would have an underdetermined system.

I got confused with the signs in the dot product. Is there a trick?

The trick is to be methodical. Write out the components clearly:

v_{L} = ⟨ - 2, d, - 4 ⟩

and

v_{1} = ⟨ 1, a, b ⟩

. The dot product is

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

. Always write the components in a column if you are prone to sign errors. It helps keep the 'x', 'y', and 'z' components aligned.

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Visualized Solution

Visualizing the 3D Geometry

Extracting Direction Ratios

Dot Product Condition for $L ⊥ L_{1}$

First Linear Equation

Dot Product Condition for $L ⊥ L_{2}$

Second Linear Equation

Eliminating $a d$ to find $b$

Point Satisfies the Line Equation

Equating Parts to Find $d$

Equating Parts to Find $c$

Back-Substitution for $a$

Calculating $a + b + c + d$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Master Equations

Solving for Parameters

Final Calculation

Similar Questions

The equation of the line through the point $(0, 1, 2)$ and perpendicular to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{- 2}$ is :

Let a line passing through the point $(4, 1, 0)$ intersect the line $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ at the point $A (α, β, γ)$ and the line $L_{2} : x - 6 = y = - z + 4$ at the point $B (a, b, c)$ . Then $1 α a 0 β b 1 γ c$ is equal to

If the length of the perpendicular from the point $(β, 0, β) (β \neq = 0)$ to the line $\frac{x - 1}{1} = \frac{y - 0}{0} = \frac{z + 1}{- 1}$ is $3/2$ , then $β$ is equal to :

If the distances of the point $(1, 2, a)$ from the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b}$ and $L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal, then $a + b + c$ is equal to

Let $L$ be the line of intersection of planes $r \cdot (\hat{i} - \hat{j} + 2 \hat{k}) = 2$ and $r \cdot (2 \hat{i} + \hat{j} - \hat{k}) = 2$ . If $P (α, β, γ)$ is the foot of perpendicular on $L$ from the point $(1, 2, 0)$ , then the value of $35 (α + β + γ)$ is equal to :

The equation of the line passing through $(- 4, 3, 1)$ , parallel to the plane $x + 2 y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ is:

Let P be the foot of the perpendicular from the point (1, 2, 2) on the line $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ Let the line $r = (- \hat{i} + \hat{j} - 2 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ , $λ \in R,$ intersect the line L at Q. Then $2 (P Q)^{2}$ is equal to:

If the length of the perpendicular drawn from the point $P (a, 4, 2), a > 0$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}$ is $26$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point $P$ in this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to :

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Visualized Solution

Visualizing the 3D Geometry

Extracting Direction Ratios

Dot Product Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L⊥L1​

First Linear Equation

Dot Product Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L⊥L2​

Second Linear Equation

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Point Satisfies the Line Equation

Equating Parts to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Equating Parts to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

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Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b+c+d

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Master Equations

Solving for Parameters

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the line through the point (0,1,2) and perpendicular to the line 2x−1​=3y+1​=−2z−1​ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the length of the perpendicular from the point (β,0,β)(β=0) to the line 1x−1​=0y−0​=−1z+1​ is 3/2​, then β is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the line passing through (−4,3,1), parallel to the plane x+2y−z−5=0 and intersecting the line −3x+1​=2y−3​=−1z−2​ is:

Dot Product Condition for $L ⊥ L_{1}$

Dot Product Condition for $L ⊥ L_{2}$

Eliminating $a d$ to find $b$

Equating Parts to Find $d$

Equating Parts to Find $c$

Back-Substitution for $a$

Calculating $a + b + c + d$

The equation of the line through the point $(0, 1, 2)$ and perpendicular to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{- 2}$ is :

If the length of the perpendicular from the point $(β, 0, β) (β \neq = 0)$ to the line $\frac{x - 1}{1} = \frac{y - 0}{0} = \frac{z + 1}{- 1}$ is $3/2$ , then $β$ is equal to :

The equation of the line passing through $(- 4, 3, 1)$ , parallel to the plane $x + 2 y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ is: