Q.Show that the three lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$; $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$; $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular.

Line 1: $\dfrac{x-1}{2}=\dfrac{y-4}{4}=\dfrac{z-2}{3}$, direction $(2,4,3)$.

Line 2: $\dfrac{1-x}{1}=\dfrac{y-2}{5}=\dfrac{3-z}{a}$. Rewrite $\dfrac{x-1}{-1}=\dfrac{y-2}{5}=\dfrac{z-3}{-a}$, direction $(-1,5,-a)$.

Perpendicular $\Rightarrow$ dot product $=0$:

We are given two lines in three-dimensional space. For two lines to be perpendicular, their direction vectors must have a dot product of zero. The challenge is that the equations are not in the standard symmetric form, so we first need to rewrite them properly.

1. Rewrite the first line in symmetric form

The first line is:

Notice that $\frac{1-x}{-3}$ can be rewritten as $\frac{x-1}{3}$ because multiplying numerator and denominator by $-1$ gives:

Also, $y$ can be written as $\frac{y-0}{1}$.

So the symmetric form of the first line is:

Thus, its direction vector is:

2. Rewrite the second line in symmetric form

The second line is:

We need each part to be of the form $\frac{x - x_0}{\text{direction}}$. For the first term:

For the middle term, $3 - y = -(y - 3)$, so we write it as:

The last term is already fine: $\frac{z-1}{a}$.

Thus the symmetric form is:

So the direction vector of the second line is:

3. Apply perpendicularity condition

Two lines are perpendicular if the dot product of their direction vectors is zero:

Substitute:

Compute:

4. Find $3a + 3b$

Multiply the equation $2a + 2b = 1$ by $\frac{3}{2}$:

Set up the line. Writing $\dfrac{2-y}{-1}=\dfrac{y-2}{1}$, the line is

so it passes through $B(1,2,3)$ with direction $\vec d=(2,1,2)$. A general point is $F=(1+2t,\;2+t,\;3+2t)$.

Foot of the perpendicular from $A(6,1,3)$.

Hence $F=(3,3,5)$ and

Perpendicular line.

so the direction ratios are $a=-3,\;b=2,\;c=2$ (equivalently $(3,-2,-2)$).

Direction of the first line: $\vec d_1 = 2\hat\imath + \hat\jmath + 2\hat k = (2,1,2).$

Rewrite the second line: $\vec r = (1,1,1) + q\,(1,\lambda,1)$, so $\vec d_2 = (1,\lambda,1).$

The angle between the lines is $\tfrac{\pi}{2}$, so the directions are perpendicular:

The key is to recognise that we are working in three-dimensional space. An equation in $x$, $y$, and $z$ generally describes a surface — here, a quadratic equation suggests a pair of planes if it factors into linear factors.

Factor the given expression:

This product equals zero when either factor is zero. So the locus is the set of all points $(x,y,z)$ satisfying:

Each of these is a plane. The plane $y=0$ is the $xz$-plane. The plane $x+z=0$ is a plane through the origin whose normal vector is $(1,0,1)$.

Now, are these planes perpendicular? Two planes are perpendicular if their normal vectors are perpendicular. The normal vector of $y=0$ is $(0,1,0)$. The dot product of $(0,1,0)$ and $(1,0,1)$ is $0 \cdot 1 + 1 \cdot 0 + 0 \cdot 1 = 0$. Since the dot product is zero, the normals are perpendicular, and therefore the planes are perpendicular.

Step 1 — Slope of the given line.

Step 2 — Slope of the perpendicular.

For perpendicular lines $m_1m_2=-1$ (their direction vectors are orthogonal):

Step 3 — Use the point $(2,2)$.

Point-slope form:

Step 4 — Read off the $y$-intercept.

In $y=mx+c$ the constant $c$ is the $y$-intercept (the value of $y$ at $x=0$):

ANSWER: C

Step 1 — Slope of the given line

Write $x + 6y = 5$ in slope-intercept form:

So its slope is $m_1 = -\dfrac{1}{6}$.

Step 2 — Slope of the required (perpendicular) line

For two perpendicular lines, $m_1 m_2 = -1$ (their direction vectors are orthogonal). Hence

Step 3 — Equation through $(-1,\,-3)$

Point-slope form $y - y_1 = m(x - x_1)$:

Step 4 — The $x$-intercept

The $x$-intercept is the value of $x$ where the line crosses the $x$-axis, i.e. where $y = 0$:

Check: the point $\left(-\tfrac12, 0\right)$ satisfies $y = 6x+3$, and the vector from $(-1,-3)$ to $(-\tfrac12, 0)$ is $(\tfrac12, 3)$, i.e. direction $(1,6)$ — indeed perpendicular to the given line's direction $(6,-1)$ since $1(6) + 6(-1) = 0$. ✓

ANSWER: B

The key insight here is that when two vectors are perpendicular, their dot product is zero. This simplifies the expression for the angle between their sum and one of them dramatically. The angle condition then becomes a clean equation relating only the magnitudes.

Let’s work through it.

1. Set up the angle condition. The angle $\theta$ between $\vec{a}+\vec{b}$ and $\vec{a}$ is given by the dot product formula:

We are told $\theta = 60^\circ$, so $\cos 60^\circ = \frac12$.

2. Simplify the numerator using $\vec{a}\cdot\vec{b}=0$.

The perpendicular condition eliminates the cross term — that’s the whole point.

3. Simplify the denominator. The magnitude of $\vec{a}+\vec{b}$ is:

So $|\vec{a}+\vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2}$.

4. Plug into the cosine equation.

Cancel one $|\vec{a}|$ (assuming $|\vec{a}| \neq 0$, which is reasonable since it’s a vector):

5. Solve for the ratio. Square both sides:

Cross-multiply:

Taking square roots (magnitudes are positive):

The concept here is the relationship between slopes of perpendicular lines. If two lines are perpendicular, the product of their slopes is $-1$ (provided neither is vertical). This is a fundamental geometric fact that lets us connect an unknown coordinate to a known line.

We are given a fixed line $2x - 3y + 17 = 0$ and told it is perpendicular to the line through $(7, 17)$ and $(15, \beta)$. Our job is to find $\beta$ that makes this true.

1. Find the slope of the given line. Rewrite $2x - 3y + 17 = 0$ in slope-intercept form $y = mx + c$:

So the slope of this line is $m_1 = \frac{2}{3}$.

2. Write the slope of the line through the two points. The slope formula: $m_2 = \frac{y_2 - y_1}{x_2 - x_1}$. Using $(x_1, y_1) = (7, 17)$ and $(x_2, y_2) = (15, \beta)$:

3. Apply the perpendicular condition. For perpendicular lines, $m_1 \cdot m_2 = -1$. Substitute:

Simplify:

Multiply both sides by 12:

So $\beta = 5$.