A plane which passes through the point (3, 2, 0) and the line (x-4)/1 = (y-7)/5 = (z-4)/4 is (a) x-y

Q: Why can't I just use the cross product of the direction vector and the vector from the point to the line?

Ah, the classic 'formula-first' trap! If you try to take the cross product of the direction vector of the line and the vector connecting the point to the line, you will get the zero vector. Why? Because the vectors are collinear! Always check for collinearity first.

Q: Wait, if there are infinite planes, how can there be only one correct option?

That is the beauty of multiple-choice questions! While there are infinitely many planes, only one of the provided options will satisfy the condition of containing the line. The others are simply 'distractors' that do not pass through the line.

Q: Do I really need to check both points?

Mathematically, yes. A plane is defined by three non-collinear points. A line is defined by two points. To ensure a plane contains a line, you must verify that at least two distinct points on that line lie on the plane. Checking only one point is a necessary condition, but not a sufficient one.

Question

A plane which passes through the point $(3, 2, 0)$ and the line $\frac{x-4}{1} = \frac{y-7}{5} = \frac{z-4}{4}$ is

Options:

(A) $x-y+z=1$

(B) $x+y+z=5$

(C) $x+2y-z=1$

(D) $2x-y+z=5$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the vast expanse of 3D space. You have a point $P(3, 2, 0)$ and a line $L$ defined by the symmetric equations:

$$
\frac{x-4}{1} = \frac{y-7}{5} = \frac{z-4}{4}
$$

The standard, textbook approach is to find the normal vector of the plane by taking the cross product of the line's direction vector and the vector connecting the point to the line.

However, in the world of JEE Advanced, the most dangerous assumption is that the geometry is 'standard.' If the point $P$ is actually sitting right on the line $L$, the line and the point would not define a unique plane. Instead, they would act like the spine of a book, with infinitely many planes passing through them.

### The Investigation: Testing the Reality

Let us test our hypothesis by substituting the coordinates of $P(3, 2, 0)$ into the line equation:

$$
\frac{3-4}{1} = \frac{2-7}{5} = \frac{0-4}{4}
$$

Calculating these ratios, we get:

$$
\frac{-1}{1} = -1, \quad \frac{-5}{5} = -1, \quad \frac{-4}{4} = -1
$$

The ratios are identical! This is a massive revelation: our point $P$ lies exactly on the line $L$. We are not looking for a unique plane; we are looking for one specific plane out of an infinite family that contains this line.

### The Strategy: The Power of Verification

Since we cannot derive a single equation from scratch, we must pivot. We know that the correct plane must contain the entire line $L$.

A plane contains a line if and only if it contains at least two distinct points on that line. We already have point $P(3, 2, 0)$. We need a second point.

Looking at the line equation, we can easily extract the base point $A(4, 7, 4)$. Now, we simply test the options provided.

### Final Verification

Let us check the option $x - y + z = 1$. Substituting $P(3, 2, 0)$, we get:

$$
3 - 2 + 0 = 1 \Rightarrow 1 = 1
$$

It works! Now, we must verify with point $A(4, 7, 4)$:

$$
4 - 7 + 4 = 1 \Rightarrow 1 = 1
$$

Both points satisfy the equation. Because this plane contains two distinct points of the line, it must contain the entire line itself.

The mystery is solved, not by brute-force calculation, but by conceptual insight and strategic verification. Keep this in your toolkit: **always check if your point lies on the line before you start your cross products!**

Animated Solution for Mathematics - Three Dimensional Geometry: A plane which passes through the point $(3, 2, 0)$ and the line $\frac{x - 4}{1} = \frac{y - 7}{5} = \frac{z - 4}{4}$ is

Select Answer:

Visualized Solution

Visualizing the Setup

The Standard Approach vs. The Trap

Substituting $P$ into $L$

Evaluating the Ratios

The Point Lies on the Line!

The Infinite Planes Concept

The Smart Strategy

Identifying a Second Point

Testing Option 1 with Point $P$

Testing Option 1 with Point $A$

Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Investigation

The Strategy

Final Verification

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Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which passes through the point (3,2,0) and the line 1x−4​=5y−7​=4z−4​ is