JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectSingle correctpyq

Three Dimensional Geometry: Mathematics | JEE Main

Q: Why do we scale the equation before finding ?

The problem specifically matches the equation to . If the constant term on the right is anything other than , the values of must be scaled proportionally.

If the equation of the plane containing the line

\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}-4=0=2 \mathrm{x}+\mathrm{y}-\mathrm{z}+5

and perpendicular to the plane

\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})

\mathrm{ax}+\mathrm{by}+\mathrm{cz}=4

, then

(\mathrm{a}-\mathrm{b}+\mathrm{c})

is equal to

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

20

24

22

18

Step-by-step solutionView

Correct answer

Using the family of planes $P_1 + \lambda P_2 = 0$ and the orthogonality condition with the normal vector of the given parametric plane, we find $\lambda = -1$ , yielding the plane $x - y - 4z - 9 = 0$ ; normalizing to the form $ax + by + cz = 4$ gives $a - b + c = 22$ .

Option analysis

Why each option works or fails

A ·

20

The student introduces a sign error when expanding the normal vector dot product $(1 + 2\lambda)(5) + (2 + \lambda)(-2) + (3 - \lambda)(-3) = 0$ , leading to an incorrect parameter $\lambda$ . Carefully track negative signs in the dot product: $5(1+2\lambda) - 2(2+\lambda) - 3(3-\lambda) = 0$ simplifies to $11\lambda - 8 = 0$ if miscalculated, instead of $11\lambda + 11 = 0$ .

B ·

24

The student calculates $a + b + c$ or computes $a - b + c$ before scaling the equation to match the right-hand constant of $4$ . Ensure the constant term matches the form $ax + by + cz = 4$ before reading off coefficients $a$ , $b$ , and $c$ .

C ·

22

None. The student correctly applies the family of planes, determines the normal to the parametric plane via the cross product, solves for $\lambda = -1$ , scales the plane to have a constant of $4$ , and computes $a - b + c = 22$ . Correct.

D ·

18

The student makes an arithmetic mistake when evaluating $(a - b + c)$ after scaling, or slips during the cross product of the direction vectors. Recalculate the cross product $(\hat{i} + \hat{j} + \hat{k}) \times (\hat{i} - 2\hat{j} + 3\hat{k}) = 5\hat{i} - 2\hat{j} - 3\hat{k}$ and verify each component.

Reviewed route

Solution

StepWorking

01given

Line of intersection: $P_1: x+2y+3z-4=0$ and $P_2: 2x+y-z+5=0$ . Plane is perpendicular to the plane $\vec{r} = (\hat{i}-\hat{j}) + \lambda(\hat{i}+\hat{j}+\hat{k}) + \mu(\hat{i}-2\hat{j}+3\hat{k})$ . The resulting plane equation is $ax+by+cz=4$ .

02goal

Find the coefficients $a, b, c$ when the constant term on the right-hand side is $4$ , and calculate $a - b + c$ .

03approach

Write the family of planes containing the line as $P_1 + k P_2 = 0$ . Find the normal vector $\vec{n}$ of the given perpendicular plane using the cross product of its spanning vectors. Impose the orthogonality condition $\vec{n}_{plane} \cdot \vec{n} = 0$ to determine $k$ , scale the equation to match the constant term $4$ , and compute $a-b+c$ .

04execute

The normal vector to the second plane is $\vec{n}_3 = (\hat{i}+\hat{j}+\hat{k}) \times (\hat{i}-2\hat{j}+3\hat{k}) = 5\hat{i} - 2\hat{j} - 3\hat{k}$ . The family of planes is $(1+2k)x + (2+k)y + (3-k)z + (5k-4) = 0$ . Since the planes are perpendicular: $5(1+2k) - 2(2+k) - 3(3-k) = 0 \implies 11k - 8 = 0 \implies k = \frac{8}{11}$ .

05execute

Substitute $k = \frac{8}{11}$ into the plane equation: $11(x+2y+3z-4) + 8(2x+y-z+5) = 0$ $27x + 30y + 25z - 4 = 0 \implies 27x + 30y + 25z = 4$ . Thus, $a = 27$ , $b = 30$ , $c = 25$ , and $a - b + c = 27 - 30 + 25 = 22$ .

✓verify

Check orthogonality: $(27)(5) + (30)(-2) + (25)(-3) = 135 - 60 - 75 = 0$ . The constant term is $-4$ , matching the right-hand side $4$ .

Hints that build this answer step by step

What is the normal vector $\vec{n}_2$ to the given plane $\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2\hat{j}+3\hat{k})$ ?
$\vec{n}_2 = 5\hat{i} - 2\hat{j} - 3\hat{k}$
Any plane containing the line of intersection can be written as $(x + 2y + 3z - 4) + \lambda(2x + y - z + 5) = 0$ . What equation determines $\lambda$ so that this plane is perpendicular to $\vec{n}_2 = 5\hat{i} - 2\hat{j} - 3\hat{k}$ ?
$5(1 + 2\lambda) - 2(2 + \lambda) - 3(3 - \lambda) = 0$
Solving $5(1+2\lambda) - 2(2+\lambda) - 3(3-\lambda) = 0$ gives $\lambda = -1$ . What is the resulting equation of the plane?
$-x + y + 4z - 9 = 0$
Writing $-x + y + 4z = 9$ in the required form $ax + by + cz = 4$ , what is the value of $(a - b + c)$ ?
$22$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why do we scale the equation before finding

a, b, c

The problem specifically matches the equation to $ax+by+cz=4$ . If the constant term on the right is anything other than $4$ , the values of $a, b, c$ must be scaled proportionally.

Answer

Using the family of planes $P_1 + \lambda P_2 = 0$ and the orthogonality condition with the normal vector of the given parametric plane, we find $\lambda = -1$ , yielding the plane $x - y - 4z - 9 = 0$ ; normalizing to the form $ax + by + cz = 4$ gives $a - b + c = 22$ .

Why each option works or fails

A: $20$ - The student introduces a sign error when expanding the normal vector dot product $(1 + 2\lambda)(5) + (2 + \lambda)(-2) + (3 - \lambda)(-3) = 0$ , leading to an incorrect parameter $\lambda$ . Carefully track negative signs in the dot product: $5(1+2\lambda) - 2(2+\lambda) - 3(3-\lambda) = 0$ simplifies to $11\lambda - 8 = 0$ if miscalculated, instead of $11\lambda + 11 = 0$ .
B: $24$ - The student calculates $a + b + c$ or computes $a - b + c$ before scaling the equation to match the right-hand constant of $4$ . Ensure the constant term matches the form $ax + by + cz = 4$ before reading off coefficients $a$ , $b$ , and $c$ .
C · correct: $22$ - None. The student correctly applies the family of planes, determines the normal to the parametric plane via the cross product, solves for $\lambda = -1$ , scales the plane to have a constant of $4$ , and computes $a - b + c = 22$ . Correct.
D: $18$ - The student makes an arithmetic mistake when evaluating $(a - b + c)$ after scaling, or slips during the cross product of the direction vectors. Recalculate the cross product $(\hat{i} + \hat{j} + \hat{k}) \times (\hat{i} - 2\hat{j} + 3\hat{k}) = 5\hat{i} - 2\hat{j} - 3\hat{k}$ and verify each component.

Step-by-step solution

given: Line of intersection: $P_1: x+2y+3z-4=0$ and $P_2: 2x+y-z+5=0$ . Plane is perpendicular to the plane $\vec{r} = (\hat{i}-\hat{j}) + \lambda(\hat{i}+\hat{j}+\hat{k}) + \mu(\hat{i}-2\hat{j}+3\hat{k})$ . The resulting plane equation is $ax+by+cz=4$ .
goal: Find the coefficients $a, b, c$ when the constant term on the right-hand side is $4$ , and calculate $a - b + c$ .
approach: Write the family of planes containing the line as $P_1 + k P_2 = 0$ . Find the normal vector $\vec{n}$ of the given perpendicular plane using the cross product of its spanning vectors. Impose the orthogonality condition $\vec{n}_{plane} \cdot \vec{n} = 0$ to determine $4$ , scale the equation to match the constant term $ax + by + cz = 4$ , and compute $a-b+c$ .
execute: The normal vector to the second plane is $a$ . The family of planes is $(1+2k)x + (2+k)y + (3-k)z + (5k-4) = 0$ . Since the planes are perpendicular: $5(1+2k) - 2(2+k) - 3(3-k) = 0 \implies 11k - 8 = 0 \implies k = \frac{8}{11}$ .
execute: Substitute $k = \frac{8}{11}$ into the plane equation: $11(x+2y+3z-4) + 8(2x+y-z+5) = 0$ $27x + 30y + 25z - 4 = 0 \implies 27x + 30y + 25z = 4$ . Thus, $a = 27$ , $b = 30$ , $c = 25$ , and $a - b + c = 27 - 30 + 25 = 22$ .
verify: Check orthogonality: $(27)(5) + (30)(-2) + (25)(-3) = 135 - 60 - 75 = 0$ . The constant term is $-4$ , matching the right-hand side $4$ .

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

Using the family of planes $P_1 + \lambda P_2 = 0$ and the orthogonality condition with the normal vector of the given parametric plane, we find $\lambda = -1$ , yielding the plane $x - y - 4z - 9 = 0$ ; normalizing to the form $ax + by + cz = 4$ gives $a - b + c = 22$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

We think you should solve this next

Eight persons are to be transported from city A to city B in three cars of different make…

Let $f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ and…

Let $\int x^{3} \sin x \, dx = g(x) + C$ , where $C$ is the constant of integration. If…

Students also ask

Three Dimensional Geometry: Mathematics | JEE Main

01given

02goal

03approach

04execute

05execute

✓verify

Eight persons are to be transported from city A to city B in three cars of different make…

Let f(x)+2f(1x)=x2+5f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5f(x)+2f(x1​)=x2+5 and…

Let ∫x3sin⁡x dx=g(x)+C\int x^{3} \sin x \, dx = g(x) + C∫x3sinxdx=g(x)+C, where CCC is the constant of integration. If…

Students also ask

Let $f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ and…

Let $\int x^{3} \sin x \, dx = g(x) + C$ , where $C$ is the constant of integration. If…