Let the position vectors of the points P, Q, R and S be a = i + 2j - 5k, b = 3i + 6j + 3k, c = (17/5

Q: Why can't we just use the scalar triple product to check for coplanarity?

You absolutely can! The scalar triple product $[\vec{PQ}, \vec{PR}, \vec{PS}] = 0$ is a powerful, standard tool. However, in this specific problem, we found an intersection point $M$ that lies on both $PR$ and $SQ$. If two lines intersect, they are coplanar by definition. Both methods are valid, but sometimes geometry offers a faster, more intuitive path than brute-force algebra.

Q: The vector $\vec{c}$ has fractions like $\frac{17}{5}$. Do I really have to work with these?

Yes, but don't panic! JEE examiners often include these fractions specifically to test your patience and precision. Notice how the denominator $5$ in $\vec{c}$ perfectly cancels out when you multiply by the ratio $5$ in the section formula. It's a gift, not a burden, designed to reward those who stay calm!

Q: Why is the cross product magnitude squared calculation so tedious?

It can feel tedious, but Lagrange's Identity ($|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$) is your best friend here. It saves you from calculating the cross product vector itself, which is prone to sign errors. Always look for these identities to streamline your work and minimize calculation time.

Question

Let the position vectors of the points $P, Q, R$ and $S$ be $\vec{a} = \hat{i} + 2\hat{j} - 5\hat{k}$, $\vec{b} = 3\hat{i} + 6\hat{j} + 3\hat{k}$, $\vec{c} = \frac{17}{5}\hat{i} + \frac{16}{5}\hat{j} + 7\hat{k}$ and $\vec{d} = 2\hat{i} + \hat{j} + \hat{k}$, respectively. Then which of the following statements is true?

Options:

(A) The points $P, Q, R$ and $S$ are NOT coplanar

(B) $\frac{\vec{b} + 2\vec{d}}{3}$ is the position vector of a point which divides $PR$ internally in the ratio $5 : 4$

(C) $\frac{\vec{b} + 2\vec{d}}{3}$ is the position vector of a point which divides $PR$ externally in the ratio $5 : 4$

(D) The square of the magnitude of the vector $\vec{b} 	imes \vec{d}$ is 95

Sigma Percentile · Accepted Answer

### The Geometry of Space: A Journey Through Vectors

Imagine you are standing in a three-dimensional coordinate system. You have four points, $P, Q, R,$ and $S$, floating in space, defined by their position vectors $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$.

At first glance, this looks like a standard vector problem, but beneath the surface lies a beautiful geometric harmony. Let's unravel it together.

### The Hidden Section Formula

We are given a mysterious vector $\vec{v}$ defined as:

$$\vec{v} = \frac{\vec{b} + 2\vec{d}}{3}$$

If you look closely, this isn't just a random combination of vectors. It is the classic section formula in disguise!

Recall that a point dividing a segment $SQ$ in a ratio $m:n$ has a position vector $\frac{m\vec{d} + n\vec{b}}{m+n}$. By rewriting $\vec{v}$ as:

$$\vec{v} = \frac{1 \cdot \vec{b} + 2 \cdot \vec{d}}{1 + 2}$$

We immediately see that this point $M$ divides the segment $SQ$ internally in a $1:2$ ratio. This is our first anchor point in the problem.

### The Test of Precision

Now, let's tackle Option B. It claims that this same point $M$ divides the segment $PR$ in a $5:4$ ratio.

To verify this, we must apply the section formula to $P$ and $R$:

$$\vec{r}_{PR} = \frac{5\vec{c} + 4\vec{a}}{5 + 4}$$

Here is where the "scary" fractions appear. We have $\vec{c} = \frac{17}{5}\hat{i} + \frac{16}{5}\hat{j} + 7\hat{k}$.

But wait—look at the ratio $5$. When we multiply $5\vec{c}$, the denominator $5$ vanishes instantly! This is the elegance of JEE problems; they test your ability to look past the initial complexity.

After substituting $\vec{a} = \hat{i} + 2\hat{j} - 5\hat{k}$ and performing the arithmetic, we find:

$$\vec{r}_{PR} = \frac{7\hat{i} + 8\hat{j} + 5\hat{k}}{3}$$

It matches our vector $\vec{v}$ perfectly! **Option B is confirmed.**

### The Power of Lagrange's Identity

Finally, let's address Option D, which asks for the square of the magnitude of the cross product $|\vec{b} 	imes \vec{d}|^2$.

Instead of calculating the cross product vector directly—which is a recipe for sign errors—we use Lagrange's Identity:

$$|\vec{b} 	imes \vec{d}|^2 = |\vec{b}|^2 |\vec{d}|^2 - (\vec{b} \cdot \vec{d})^2$$

Calculating $|\vec{b}|^2 = 54$, $|\vec{d}|^2 = 6$, and $\vec{b} \cdot \vec{d} = 15$ is straightforward. Plugging these in, we get:

$$54 	imes 6 - 15^2 = 324 - 225 = 99$$

Since $99 
eq 95$, **Option D is incorrect.**

### The Final Insight

We have discovered that point $M$ lies on both line segments $PR$ and $SQ$. Because these two lines intersect at $M$, they must lie in the same plane.

Therefore, the points $P, Q, R,$ and $S$ are coplanar. This invalidates Option A.

Through this journey, we haven't just solved a problem; we've seen how vectors, section formulas, and geometric identities weave together to describe the structure of space. Keep practicing, and soon, you won't just see equations—you'll see the geometry itself.

Select Answer:

Visualized Solution

Given Position Vectors

Analyzing Options B and C

Substituting $b$ and $d$

Simplifying $v$

Geometric Meaning of $v$

Testing Option B

Substituting $c$ and $a$

Multiplying Scalars

Adding Components

Simplifying the Fraction

Conclusion for Option B

Checking Option D

Calculating Magnitudes and Dot Product

Applying Lagrange's Identity

The Hidden Geometric Truth

The Sigma Insight: Addition of Vectors

The Geometry of Space

The Hidden Section Formula

The Test of Precision

The Power of Lagrange's Identity

The Final Insight

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