Q.Answer the following as true or false.

The key idea: three points are collinear if and only if the displacement vectors between any two pairs are parallel — that is, one is a scalar multiple of the other. This gives us two equations in $x$ and $y$, which we solve directly.

Let the points be:

$A: x\vec{i} + 2\vec{j} + y\vec{k}$

$B: \vec{i} - 2\vec{j} + 2\vec{k}$

$C: 2\vec{i} + 3\vec{j} - \vec{k}$

For collinearity, vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ (or $\overrightarrow{AC}$) must be parallel.

1. Compute $\overrightarrow{AB} = B - A$:

2. Compute $\overrightarrow{BC} = C - B$:

3. Since $\overrightarrow{AB}$ is parallel to $\overrightarrow{BC}$, there exists a scalar $\lambda$ such that:

That is:

4. Equate the components:

   • $\vec{i}$: $1 - x = \lambda$
   • $\vec{j}$: $-4 = 5\lambda$
   • $\vec{k}$: $2 - y = -3\lambda$

   From the $\vec{j}$ equation: $\lambda = -\frac{4}{5}$.

5. Substitute $\lambda$ into the $\vec{i}$ equation:

6. Substitute $\lambda$ into the $\vec{k}$ equation: $$$$

The key idea is that if points A, B, C are collinear, then B lies on the line through A and C. That means the position vectors satisfy the section formula: there exists a ratio $\lambda : \mu$ such that

where the sign of $\mu$ and $\lambda$ tells us whether the division is internal or external. We can find this ratio by comparing coefficients of $\vec{i}, \vec{j}, \vec{k}$.

Let’s work through it step by step.

1. Write the given vectors clearly

   $\vec{a} = \hat{i} - 6\hat{j} + 9\hat{k}$

   $\vec{b} = \hat{i} + 3\hat{j} + 5\hat{k}$

   $\vec{c} = 2\hat{i} + \beta\hat{j} + 7\hat{k}$

   We don’t yet know $\beta$, but collinearity will determine it.

2. Apply the section formula

   Suppose B divides AC in the ratio $m:n$. Then

(Here $m$ corresponds to the segment from A to B, $n$ from B to C; the formula is symmetric — just be consistent.)

3. Equate the coefficients of $\hat{i}$ From the $\hat{i}$ components:

Multiply: $m + n = n + 2m$

Cancel $n$: $m = 2m$ → $m = 0$? That can’t be right — let’s check carefully.

Actually: $m + n = n + 2m$ gives $m + n - n = 2m$, so $m = 2m$, hence $m = 0$. That would mean B coincides with A, which is false. So something is off — we must have used the wrong assignment of $m$ and $n$.

Let’s instead write: Let B divide AC in the ratio $k:1$ (i.e., $AB:BC = k:1$). Then

because the weight of A is the segment opposite to it (BC) and the weight of C is the segment opposite to it (AB). Check: if $k=0$, B = A; if $k \to \infty$, B = C. That’s correct.

4. Equate $\hat{i}$ components with this form

So $1 + k = 1 + 2k$ → $1 + k = 1 + 2k$ → $k = 2k$ → $k = 0$. Again zero? That suggests the $\hat{i}$ components are inconsistent unless the points are not collinear with this assignment — but they are given collinear. The only resolution: the division is external, so the formula uses a negative sign.

For external division in ratio $k:1$, we have

(or the other sign, depending on which side B lies). Let’s try:

Multiply: $1 - k = 1 - 2k$ → $-k = -2k$ → $k = 2k$ → $k=0$ again? That’s still problematic. Let’s instead use the symmetric form:

and allow $\mu$ and $\lambda$ to have opposite signs for external division.

5. Use the $\hat{j}$ component to find the ratio From $\hat{j}$:

And from $\hat{k}$:

The $\hat{k}$ equation doesn’t involve $\beta$, so solve it first.

From $\hat{k}$: $5(\mu + \lambda) = 9\mu + 7\lambda$

→ $5\mu + 5\lambda = 9\mu + 7\lambda$

→ $5\lambda - 7\lambda = 9\mu - 5\mu$

→ $-2\lambda = 4\mu$

→ $\lambda = -2\mu$

So $\mu$ and $\lambda$ have opposite signs — that means external division. The ratio $\lambda : \mu = -2\mu : \mu = -2:1$, or equivalently $2:1$ externally with B dividing AC. But careful: the ratio in which B divides AC is $|\lambda| : |\mu| = 2:1$, and since signs differ, it’s external.

6. Verify with $\hat{i}$ component Using $\lambda = -2\mu$:

Check $\hat{i}$: $2(2) - 1 = 3$, but $\vec{b}$ has $\hat{i}$ component 1. That’s a mismatch — so our assignment of which vector corresponds to which coefficient might be reversed.

Let’s instead set $\vec{b} = \frac{\lambda \vec{a} + \mu \vec{c}}{\lambda + \mu}$ and solve again. From $\hat{k}$:

$5(\lambda + \mu) = 9\lambda + 7\mu$

→ $5\lambda + 5\mu = 9\lambda + 7\mu$

→ $5\mu - 7\mu = 9\lambda - 5\lambda$

→ $-2\mu = 4\lambda$

→ $\mu = -2\lambda$

So $\mu : \lambda = -2:1$, meaning the ratio $|\mu|:|\lambda| = 2:1$ externally. Now check $\hat{i}$:

$\vec{b} = \frac{\lambda \vec{a} + (-2\lambda) \vec{c}}{\lambda - 2\lambda} = \frac{\lambda(\vec{a} - 2\vec{c})}{-\lambda} = 2\vec{c} - \vec{a}$

$\hat{i}$: $2(2) - 1 = 3$, but we need 1. So this still fails — meaning the $\hat{i}$ components are inconsistent unless the points are not collinear as given? But they are. The only possibility: the $\hat{i}$ component of $\vec{a}$ or $\vec{c}$ is miswritten? No, the problem states them as given. Let’s re-check the original: $\vec{a} = i - 6j + 9k$, $\vec{b} = i + 3j + 5k$, $\vec{c} = 2i + \beta j + 7k$. For collinearity, the vectors $\vec{b} - \vec{a}$ and $\vec{c} - \vec{a}$ must be parallel.

$\vec{b} - \vec{a} = (1-1)i + (3+6)j + (5-9)k = 0i + 9j -4k$

$\vec{c} - \vec{a} = (2-1)i + (\beta+6)j + (7-9)k = 1i + (\beta+6)j -2k$

For parallelism: $\frac{0}{1} = \frac{9}{\beta+6} = \frac{-4}{-2}$ …

Reference vectors. With $P_1=(1,-2,3),\ P_2=(2,3,-4),\ P_3=(-3,1,-5),\ P_4=(a,-2,4)$:

Coplanarity requires the scalar triple product to vanish:

Expand along the third row (it has a zero): …

Let $P=(\alpha,4,2)$, $Q=(6,\beta,-1)$, $R=(8,-\beta,-7)$.

For collinearity $\vec{PQ}\parallel\vec{QR}$, so the ratios are equal. Using the $z$-components, $\dfrac{-3}{-6}=\dfrac12$ is the common ratio. …

We are given the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Its parametric coordinates are $(a \sec \theta, b \tan \theta)$. So points A and B are $A(a \sec \theta_1, b \tan \theta_1)$ and $B(a \sec \theta_2, b \tan \theta_2)$. The focus S of the hyperbola is at $(ae, 0)$, where $e = \sqrt{1 + \frac{b^2}{a^2}}$, so $ae = \sqrt{a^2 + b^2}$.

The condition that A, S, B are collinear means that the slope from A to S equals the slope from S to B. This gives an equation linking $\theta_1$ and $\theta_2$. We then manipulate that equation to match the given form $a \cos\left(\frac{\theta_1 + \theta_2}{2}\right) = k \cos\left(\frac{\theta_1 - \theta_2}{2}\right)$ and solve for $k$.

1. Write the collinearity condition. Points: $A(a\sec\theta_1, b\tan\theta_1)$, $S(\sqrt{a^2+b^2}, 0)$, $B(a\sec\theta_2, b\tan\theta_2)$. Slope AS = slope SB:

Notice the denominators are opposites in a sense; cross-multiplying gives:

Cancel $b$ (assuming $b \neq 0$):

2. Rearrange to isolate terms. Expand both sides:

Bring terms with $\sqrt{a^2+b^2}$ together and terms with $a$ together:

3. Use trigonometric identities. Recall:

Also:

Interestingly, both sides have the same numerator and denominator! So the equation becomes:

4. Simplify. Since $\theta_1 \neq \theta_2$ (otherwise A and B coincide), $\sin(\theta_1 - \theta_2) \neq 0$, and $\cos\theta_1\cos\theta_2 \neq 0$ (points are on the hyperbola, so secants are defined). Cancel the common factor:

This seems to force a relation between $a$ and $b$ that is not generally true. What went wrong?

Pitfall: We assumed the denominators in the slope equality are non-zero and directly cross-multiplied, but we missed that the signs might flip differently. Let's re-derive carefully.

5. Use the collinearity determinant. Points $A(x_1,y_1)$, $S(x_0,0)$, $B(x_2,y_2)$ are collinear if:

Expanding:

So:

Rearranging:

6. Substitute parametric forms. Here $x_0 = \sqrt{a^2+b^2}$, $x_1 = a\sec\theta_1$, $y_1 = b\tan\theta_1$, $x_2 = a\sec\theta_2$, $y_2 = b\tan\theta_2$. Then:

Cancel $b$:

7. Simplify the right-hand side. Write in sines and cosines:

The left-hand side:

So both sides are exactly the same factor times $\frac{\sin(\theta_2 - \theta_1)}{\cos\theta_1\cos\theta_2}$. Cancel this common factor (non-zero):

$\sqrt{a^2+b^2} = a.$ …