If the angle between the lines, x/2 = y/2 = z/1 and (5-x)/-2 = (7y-14)/p = (z-3)/4 is cos⁻¹(2/3), th

Q: Wait, why do we have to standardize the line equation? Can't we just take the denominators as they are?

That is a classic trap! In the equation $\frac{5-x}{-2}$, the coefficient of $x$ is $-1$. The standard form requires the coefficient of the variable to be $+1$. If you take the denominator as $-2$ directly, you are essentially using the wrong direction vector. By dividing the numerator and denominator by $-1$, you get $\frac{x-5}{2}$, which reveals the true direction ratio is $2$, not $-2$. Always standardize first to avoid this sign error.

Q: Why is there an absolute value in the dot product formula? Does it matter if the angle is acute or obtuse?

Great question! The angle between two lines is conventionally defined as the acute angle. The dot product $\vec{a} \cdot \vec{b}$ can be negative, which would give a negative cosine value, implying an obtuse angle. By taking the absolute value, we ensure that we are calculating the acute angle between the lines. In this problem, since we are given $\cos \theta = \frac{2}{3}$, which is positive, the absolute value is essential to maintain consistency with the given value.

Q: I got a quadratic equation in $p$ after squaring, but then the $p^2$ terms cancelled out. Did I do something wrong?

Not at all! That is actually a sign that you are on the right track. In many JEE problems, complex-looking terms often cancel out, revealing the elegance of the underlying geometry. If the $p^2$ terms cancel, it simply means the relationship between the parameter $p$ and the angle is linear, not quadratic. Trust your algebra—if the cancellation is mathematically sound, you have likely solved it correctly.

Question

If the angle between the lines, $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{5-x}{-2} = \frac{7y-14}{p} = \frac{z-3}{4}$ is $\cos^{-1}\left(\frac{2}{3}\right)$, then $p$ is equal to :

Options:

(A) $-\frac{4}{7}$

(B) $\frac{7}{2}$

(C) $-\frac{7}{4}$

(D) $\frac{2}{7}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The first step is the most critical. Many students rush into the dot product formula without checking the form of the lines.

Look at $L_1$: 
$$\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$$
This is already in the standard symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. We can immediately identify the direction vector $\vec{a} = 2\hat{i} + 2\hat{j} + \hat{k}$.

Now, consider $L_2$: 
$$\frac{5-x}{-2} = \frac{7y-14}{p} = \frac{z-3}{4}$$
This is the trap. The coefficients of $x$ and $y$ are not $+1$. We must manipulate this to reach the standard form.

For the first term, we divide the numerator and denominator by $-1$ to get $\frac{x-5}{2}$. For the second term, we divide by $7$ to get $\frac{y-2}{p/7}$.

Now, the line is in standard form. The direction vector $\vec{b}$ is $2\hat{i} + \frac{p}{7}\hat{j} + 4\hat{k}$.

### The Bridge of Geometry

Now that we have our vectors $\vec{a}$ and $\vec{b}$, we use the bridge between algebra and geometry: the dot product formula:
$$\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$$
We are given $\cos 	heta = \frac{2}{3}$. This is our anchor.

Let us calculate the dot product $\vec{a} \cdot \vec{b}$:
$$\vec{a} \cdot \vec{b} = (2)(2) + (2)\left(\frac{p}{7}ight) + (1)(4) = 8 + \frac{2p}{7}$$

Next, we calculate the magnitudes:
$$|\vec{a}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3$$
$$|\vec{b}| = \sqrt{2^2 + \left(\frac{p}{7}ight)^2 + 4^2} = \sqrt{20 + \frac{p^2}{49}}$$

### The Algebraic Dance

Now, we assemble the pieces:
$$\frac{2}{3} = \frac{|8 + \frac{2p}{7}|}{3 \sqrt{20 + \frac{p^2}{49}}}$$

Notice the $3$ in the denominator on the right and the $3$ in the denominator on the left; they cancel out beautifully. We are left with:
$$2 = \frac{|8 + \frac{2p}{7}|}{\sqrt{20 + \frac{p^2}{49}}}$$

To solve for $p$, we square both sides. This removes the absolute value and the square root:
$$4 = \frac{(8 + \frac{2p}{7})^2}{20 + \frac{p^2}{49}}$$

Cross-multiply to obtain:
$$4\left(20 + \frac{p^2}{49}ight) = \left(8 + \frac{2p}{7}ight)^2$$

Expanding this, we get:
$$80 + \frac{4p^2}{49} = 64 + \frac{4p^2}{49} + 2(8)\left(\frac{2p}{7}ight)$$

Here is the moment of magic: the $\frac{4p^2}{49}$ terms appear on both sides and vanish. We are left with a simple linear equation:
$$80 = 64 + \frac{32p}{7}$$

Subtracting $64$ from $80$ gives $16 = \frac{32p}{7}$. Finally, dividing by $16$ gives $1 = \frac{2p}{7}$, which leads us to the final result:
**$$p = \frac{7}{2}$$**

Animated Solution for Mathematics - Three Dimensional Geometry: If the angle between the lines, $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{5 - x}{- 2} = \frac{7 y - 14}{p} = \frac{z - 3}{4}$ is $cos^{- 1} (\frac{2}{3})$ , then $p$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Problem

Standardizing Line $L_{1}$

Standardizing Line $L_{2}$

Direction Vector of $L_{2}$

The Angle Formula

Calculating Dot Product $a \cdot b$

Calculating Magnitudes

Substituting in Formula

Simplifying the Equation

Squaring Both Sides

Expanding and Solving

Final Calculation for $p$

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Bridge of Geometry

The Algebraic Dance

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Standardizing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Standardizing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

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