If the distance of the point P(43, α, β), β

Q: Why can't we just use the perpendicular distance formula?

The problem explicitly states the distance is measured 'along a line with direction ratios 3, -1, 0'. This is a projection problem, not a shortest-distance problem. Using the perpendicular distance formula would assume the path is orthogonal to the line, which is not the case here.

Q: Why did we set the third ratio to zero?

In the proportionality $\frac{x_Q - x_P}{3} = \frac{y_Q - y_P}{-1} = \frac{z_Q - z_P}{0} = \lambda$, the third term has a zero denominator. For this to be a valid, finite ratio, the numerator $z_Q - z_P$ must be zero. This is a fundamental property of direction ratios.

Q: Why do we reject the positive $\alpha$ case?

We found two possible values for $\alpha$ (13 and -13). When we tested $\alpha = 13$, we calculated $\beta = 116$. The problem explicitly states the constraint $\beta < 0$. Since 116 is not less than 0, we must reject this case.

Question

If the distance of the point $P(43, \alpha, \beta), \beta < 0$, from the line $\vec{r} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k}), \mu \in \mathbb{R}$ along a line with direction ratios $3, -1, 0$ is $13\sqrt{10}$, then $\alpha^2 + \beta^2$ is equal to ____

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the three-dimensional realm! Today, we are going to tackle a problem that might seem like a standard distance calculation, but it hides a beautiful geometric secret.

Imagine a point $P(43, \alpha, \beta)$ floating in space. We are given a line $L_1$ defined by the vector equation:

$$\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$$

Our goal is to measure the distance from $P$ to $L_1$ along a specific direction vector $\vec{d} = (3, -1, 0)$. This is a crucial distinction; we are not dropping a perpendicular, but traveling along a specific path to hit the line at a point $Q$.

Let's keep our eyes on the constraint $\beta < 0$; it will be our compass when we reach the final, decisive moment.

### Defining the Intersection Point $Q$

To find the distance, we first need to identify the point $Q$ where our path intersects the line $L_1$. Since $Q$ lies on $L_1$, its coordinates must satisfy the line's equation.

By reading the coefficients from $\vec{r} = (4 + 2\mu)\hat{i} + 0\hat{j} + (-1 + 3\mu)\hat{k}$, we can write the coordinates of $Q$ as:

$$Q = (4 + 2\mu, 0, -1 + 3\mu)$$

Now, we define the vector $\vec{PQ}$ by subtracting the coordinates of $P$ from $Q$:

$$\vec{PQ} = ((4 + 2\mu) - 43, 0 - \alpha, (-1 + 3\mu) - \beta)$$

Simplifying this, we get:

$$\vec{PQ} = (2\mu - 39, -\alpha, 3\mu - \beta - 1)$$

### The Parallelism Constraint

We are told the distance is measured along a line with direction ratios $(3, -1, 0)$. This means our vector $\vec{PQ}$ must be parallel to the direction vector $\vec{d} = (3, -1, 0)$.

Mathematically, this implies that the direction ratios of $\vec{PQ}$ must be proportional to the direction ratios of $\vec{d}$. We set up the proportionality:

$$\frac{2\mu - 39}{3} = \frac{-\alpha}{-1} = \frac{3\mu - \beta - 1}{0} = \lambda$$

Look closely at that third ratio. A zero in the denominator is not a sign of failure; it is a sign of a constraint! For the ratio to be finite, the numerator must be zero.

Thus, $3\mu - \beta - 1 = 0$, or $\beta = 3\mu - 1$. This is a powerful simplification.

### The Algebraic Dance

Now, let's use the distance information. We are given $PQ = 13\sqrt{10}$. The distance $PQ$ is the magnitude of the vector $\vec{PQ}$.

Using our proportionality, we can express the components of $\vec{PQ}$ in terms of $\alpha$. Since $\frac{2\mu - 39}{3} = \lambda$ and $\frac{-\alpha}{-1} = \lambda$, we have $2\mu - 39 = 3\alpha$.

Substituting these into the distance formula:

$$PQ = \sqrt{(3\alpha)^2 + (-\alpha)^2 + 0^2} = \sqrt{9\alpha^2 + \alpha^2} = \sqrt{10\alpha^2} = |\alpha|\sqrt{10}$$

Equating this to the given distance $13\sqrt{10}$, we find $|\alpha| = 13$. This gives us two potential paths: $\alpha = 13$ or $\alpha = -13$.

### The Final Verification

We must now test these cases against our constraint $\beta < 0$.

If $\alpha = 13$, then $2\mu - 39 = 3(13) = 39$, which means $2\mu = 78$, so $\mu = 39$. Then $\beta = 3(39) - 1 = 116$. This violates $\beta < 0$.

If $\alpha = -13$, then $2\mu - 39 = 3(-13) = -39$, which means $2\mu = 0$, so $\mu = 0$. Then $\beta = 3(0) - 1 = -1$. This satisfies $\beta < 0$.

With $\alpha = -13$ and $\beta = -1$, the final value is:

$$\alpha^2 + \beta^2 = (-13)^2 + (-1)^2 = 169 + 1 = \mathbf{170}$$

Animated Solution for Mathematics - Three Dimensional Geometry: If the distance of the point $P (43, α, β), β < 0$ , from the line $r = 4 \hat{i} - \hat{k} + μ (2 \hat{i} + 3 \hat{k}), μ \in R$ along a line with direction ratios $3, - 1, 0$ is $1310$ , then $α^{2} + β^{2}$ is equal to ____

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Direction and Distance

Defining the Intersection Point $Q$

Formulating Vector $P Q$

Applying the Parallel Condition

Extracting Equations from Proportions

Expressing Distance $P Q$ in terms of $α$

Solving for $α$

Testing Case 1 ( $α = 13$ )

Testing Case 2 ( $α = - 13$ )

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Defining the Intersection Point $Q$

The Parallelism Constraint

The Algebraic Dance

The Final Verification

Similar Questions

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:

Let the point, on the line passing through the points $P (1, - 2, 3)$ and $Q (5, - 4, 7)$ , farther from the origin and at distance of 9 units from the point $P$ , be $(α, β, γ)$ . Then $α^{2} + β^{2} + γ^{2}$ is equal to :

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The distance of the point having position vector $- \hat{i} + 2 \hat{j} + 6 \hat{k}$ from the straight line passing through the point $(2, 3, - 4)$ and parallel to the vector, $6 \hat{i} + 3 \hat{j} - 4 \hat{k}$ is :

Let $P (- 2, - 1, 1)$ and $Q (\frac{56}{17}, \frac{43}{17}, \frac{111}{17})$ be the vertices of the rhombus $P R QS$ . If the direction ratios of the diagonal $R S$ are $α, - 1, β$ , where both $α$ and $β$ are integers of minimum absolute values, then $α^{2} + β^{2}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Direction and Distance

Defining the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Formulating Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ​

Applying the Parallel Condition

Extracting Equations from Proportions

Expressing Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Testing Case 1 ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=13)

Testing Case 2 ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=−13)

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Defining the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

The Parallelism Constraint

The Algebraic Dance

The Final Verification

Similar Questions

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Formulating Vector $P Q$

Expressing Distance $P Q$ in terms of $α$

Solving for $α$

Testing Case 1 ( $α = 13$ )

Testing Case 2 ( $α = - 13$ )

Defining the Intersection Point $Q$

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:

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