Q.Find the values of $p$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angles.

Concept: Perpendicular Vectors Condition — two lines are perpendicular when the dot product of their direction vectors is zero.

Step 1: Write direction vectors in standard form.

First line:

$\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$

Rewrite as $\frac{x-1}{-3} = \frac{y-2}{\frac{2p}{7}} = \frac{z-3}{2}$

So direction vector $\vec{d_1} = \left(-3,\ \frac{2p}{7},\ 2\right)$.

Second line:

$\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$

Rewrite as $\frac{x-1}{-\frac{3p}{7}} = \frac{y-5}{1} = \frac{z-6}{-5}$

So direction vector $\vec{d_2} = \left(-\frac{3p}{7},\ 1,\ -5\right)$.

Step 2: Apply perpendicular condition. …

We need two lines to be perpendicular. In 3D geometry, two lines are at right angles when their direction vectors are perpendicular — meaning their dot product equals zero. The key is to first extract the direction vectors from the given symmetric equations, then set up and solve that dot product equation.

Let’s rewrite each line in standard symmetric form: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $(a, b, c)$ is the direction vector.

1. First line:

Given: $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$

Rewrite $\frac{1-x}{3}$ as $\frac{x-1}{-3}$ (multiply numerator and denominator by $-1$).

For the $y$-term: $\frac{7y-14}{2p} = \frac{7(y-2)}{2p} = \frac{y-2}{\frac{2p}{7}}$.

The $z$-term is already fine: $\frac{z-3}{2}$.

So the first line in standard form is:

Direction vector $\vec{d_1} = \left( -3,\ \frac{2p}{7},\ 2 \right)$.

2. Second line:

Given: $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$

Rewrite $\frac{7-7x}{3p} = \frac{7(1-x)}{3p} = \frac{1-x}{\frac{3p}{7}} = \frac{x-1}{-\frac{3p}{7}}$.

For $z$: $\frac{6-z}{5} = \frac{z-6}{-5}$.

So the second line in standard form is:

Direction vector $\vec{d_2} = \left( -\frac{3p}{7},\ 1,\ -5 \right)$.

3. Perpendicular condition:

Two vectors are perpendicular iff their dot product is zero:

Compute:

Simplify term by term: …

Method: Solve for an unknown that makes two lines perpendicular

When a line contains an unknown (here $p$) and a right-angle condition is imposed, write both direction vectors, set their dot product to zero, and solve the resulting equation for the unknown.

Steps

Step 1: Rewrite each line in clean standard form $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$. This is where errors hide:

• a numerator like $1-x$ must become $-(x-1)$, flipping the denominator's sign;
• a numerator like $7y-14=7(y-2)$ carries a coefficient, so the effective denominator is $\dfrac{2p}{7}$, not $2p$.

Step 2: Read the direction vectors $\vec d_1,\vec d_2$ from the tidied denominators. …

Common Mistakes

Mistake 1: Reading direction ratios straight from the raw fractions without tidying.

Why it's wrong: $\frac{1-x}{3}$ hides a sign ($=\frac{x-1}{-3}$, direction $-3$), and $\frac{7y-14}{2p}=\frac{y-2}{2p/7}$ hides a coefficient (direction $\frac{2p}{7}$, not $2p$). Correct approach: rewrite every term as $\frac{x-x_1}{a}$ first, then read $\vec d_1=(-3,\frac{2p}{7},2)$, $\vec d_2=(-\frac{3p}{7},1,-5)$.

Mistake 2: Forgetting the $7$ inside $7y-14$ and $7-7x$. …