Key Concepts and Formulas

• Standard Form of a Line in 3D: A line in three-dimensional space is typically represented in its symmetric form as: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ Here, $(x_1, y_1, z_1)$ is a point through which the line passes, and $(a, b, c)$ are the direction ratios (DRs) of the line. These direction ratios are proportional to the components of a vector parallel to the line.
• Angle Between Two Lines: If two lines have direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ respectively, the cosine of the acute angle $\theta$ between them is given by the formula: $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$ The absolute value ensures that we calculate the acute angle, which is the standard interpretation unless otherwise specified.

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Step-by-Step Solution

Step 1: Identify Direction Ratios of the First Line

The first line is given by the equation: $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ Why: This equation is already in the standard symmetric form, $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $(x_1, y_1, z_1) = (0, 0, 0)$. By direct comparison, we can identify the direction ratios for the first line: $(a_1, b_1, c_1) = (2, 2, 1)$

Step 2: Transform the Second Line into Standard Form and Identify its Direction Ratios

The second line is given by the equation: $\frac{5 - x}{-2} = \frac{7y - 14}{p} = \frac{z - 3}{4}$ Why: Before we can identify the direction ratios $(a_2, b_2, c_2)$, we must convert this equation into the standard symmetric form. This specifically requires the coefficients of $x, y, z$ in the numerators to be exactly $+1$. Let's transform each part:

1. For the $x$-term: The numerator is $5 - x$. To match the standard form $(x - x_1)$, we factor out $-1$: $5 - x = -(x - 5)$. So, the term becomes $\frac{-(x - 5)}{-2}$. We can cancel the negative signs: $\frac{x - 5}{2}$.
2. For the $y$-term: The numerator is $7y - 14$. To make the coefficient of $y$ equal to $1$, we factor out $7$: $7y - 14 = 7(y - 2)$. So, the term becomes $\frac{7(y - 2)}{p}$. To isolate $(y - 2)$ in the numerator and move the constant factor to the denominator, we divide the denominator by $7$: $\frac{y - 2}{p/7}$.
3. For the $z$-term: The $z$-term $\frac{z - 3}{4}$ is already in the correct standard form, as the coefficient of $z$ is $+1$.

Now, substituting these transformed terms back into the equation for the second line, we get its standard symmetric form: $\frac{x - 5}{2} = \frac{y - 2}{p/7} = \frac{z - 3}{4}$ From this standard form, we can identify the direction ratios for the second line: $(a_2, b_2, c_2) = \left(2, \frac{p}{7}, 4\right)$

Step 3: Apply the Angle Formula with Given Information

We are given that the angle between the lines is $\theta = \cos^{-1}\left(\frac{2}{3}\right)$. This implies $\cos \theta = \frac{2}{3}$.

Now, we substitute the direction ratios $(a_1, b_1, c_1) = (2, 2, 1)$ and $(a_2, b_2, c_2) = \left(2, \frac{p}{7}, 4\right)$ and the value of $\cos \theta$ into the angle formula: $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$ $\frac{2}{3} = \left| \frac{(2)(2) + (2)\left(\frac{p}{7}\right) + (1)(4)}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{2^2 + \left(\frac{p}{7}\right)^2 + 4^2}} \right|$ Let's simplify the numerator and the terms under the square roots:

• Numerator: $4 + \frac{2p}{7} + 4 = 8 + \frac{2p}{7}$.
• First square root term: $\sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
• Second square root term: $\sqrt{2^2 + \left(\frac{p}{7}\right)^2 + 4^2} = \sqrt{4 + \frac{p^2}{49} + 16} = \sqrt{20 + \frac{p^2}{49}}$.

Substitute these simplified expressions back into the equation: $\frac{2}{3} = \left| \frac{8 + \frac{2p}{7}}{3 \sqrt{20 + \frac{p^2}{49}}} \right|$ Why: To simplify, we can cancel the $3$ in the denominator on the right side with the $3$ on the left side. $2 = \left| \frac{8 + \frac{2p}{7}}{\sqrt{20 + \frac{p^2}{49}}} \right|$

Step 4: Solve for p

Why: To solve for $p$, we need to eliminate the absolute value and the square root. Squaring both sides is the most direct method to achieve this. $2^2 = \frac{\left(8 + \frac{2p}{7}\right)^2}{\left(\sqrt{20 + \frac{p^2}{49}}\right)^2}$ $4 = \frac{\left(8 + \frac{2p}{7}\right)^2}{20 + \frac{p^2}{49}}$ Now, cross-multiply: $4 \left(20 + \frac{p^2}{49}\right) = \left(8 + \frac{2p}{7}\right)^2$ Expand both sides of the equation:

• Left Hand Side (LHS): $4 \times 20 + 4 \times \frac{p^2}{49} = 80 + \frac{4p^2}{49}$.
• Right Hand Side (RHS): Using the algebraic identity $(A+B)^2 = A^2 + 2AB + B^2$, where $A=8$ and $B=\frac{2p}{7}$: $\left(8 + \frac{2p}{7}\right)^2 = 8^2 + 2 \times 8 \times \frac{2p}{7} + \left(\frac{2p}{7}\right)^2$ $= 64 + \frac{32p}{7} + \frac{4p^2}{49}$ Equate LHS and RHS: $80 + \frac{4p^2}{49} = 64 + \frac{32p}{7} + \frac{4p^2}{49}$ Why: Notice that the term $\frac{4p^2}{49}$ appears on both sides of the equation. We can cancel it out, simplifying the equation significantly. $80 = 64 + \frac{32p}{7}$ Now, solve for $p$: $80 - 64 = \frac{32p}{7}$ $16 = \frac{32p}{7}$ Multiply both sides by $7$: $16 \times 7 = 32p$ $112 = 32p$ Divide by $32$: $p = \frac{112}{32}$ To simplify the fraction, divide both numerator and denominator by common factors (e.g., 16): $p = \frac{112 \div 16}{32 \div 16} = \frac{7}{2}$ Verification (Optional): We assumed $8 + \frac{2p}{7}$ was positive when we removed the absolute value by squaring. Let's check with $p = \frac{7}{2}$: $8 + \frac{2}{7} \left(\frac{7}{2}\right) = 8 + 1 = 9$. Since $9 > 0$, our assumption was consistent, and $p = \frac{7}{2}$ is the valid solution.

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Common Mistakes & Tips

• Non-Standard Form: The most common mistake is to directly pick the denominators as direction ratios without first ensuring that the coefficients of $x, y, z$ in the numerators are $+1$. Always transform the equation to the standard symmetric form.
• Algebraic Errors: Be careful with fractions and squaring binomials. Double-check calculations, especially when dealing with variables in denominators.
• Absolute Value: Remember the absolute value in the angle formula $\cos \theta = \left| \dots \right|$. This ensures you find the acute angle. Squaring both sides correctly handles the absolute value.

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Summary

To find the value of $p$, we first extracted the direction ratios of the first line directly from its standard form. For the second line, we carefully transformed its equation into the standard symmetric form by adjusting the numerators to have coefficients of $+1$ for $x, y, z$, which in turn modified its denominators to represent the correct direction ratios. Then, we applied the formula for the cosine of the angle between two lines, substituted the given angle, and solved the resulting algebraic equation for $p$.

The final answer is \boxed{\text{7 \over 2}}, which corresponds to option (A).