Let L be the line of intersection of planes r · (i - j + 2k) = 2 and r · (2i + j - k) = 2. If P(α, β

Q: Why can't we just use the distance formula to find the foot of the perpendicular?

The distance formula gives you the length of the perpendicular segment, but it doesn't give you the coordinates of the foot itself. To find the foot, you need to use the parametric form of the line and the dot product condition, which ensures the vector from the external point to the line is perpendicular to the line's direction.

Q: Why did we set $z=0$ to find a point on the line? Is that always valid?

The line of intersection is infinite, so it must pass through the $xy$-plane (where $z=0$) unless it is parallel to it. Setting $z=0$ is just a convenient choice to simplify the system of equations. Any point on the line will work, so this is a perfectly valid strategy.

Q: Is the order of the cross product $\vec{n}_1 \times \vec{n}_2$ important?

The order determines the direction of the vector, but not the line itself. $\vec{n}_1 \times \vec{n}_2$ and $\vec{n}_2 \times \vec{n}_1$ are just opposite vectors, meaning they define the same line. You can use either one, and the final result will be the same.

Question

Let $L$ be the line of intersection of planes $\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 2$ and $\vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) = 2$. If $P(\alpha, \beta, \gamma)$ is the foot of perpendicular on $L$ from the point $(1, 2, 0)$, then the value of $35(\alpha + \beta + \gamma)$ is equal to :

Options:

(A) 101

(B) 119

(C) 143

(D) 134

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The line $L$ is formed by the intersection of two planes, $\pi_1$ and $\pi_2$, with normal vectors $\vec{n}_1 = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{n}_2 = 2\hat{i} + \hat{j} - \hat{k}$.

Since the line $L$ lies on both planes, its direction vector $\vec{d}$ must be perpendicular to both $\vec{n}_1$ and $\vec{n}_2$. We find this direction using the cross product:

$$\vec{d} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 2 & 1 & -1 \end{vmatrix} = -\hat{i} + 5\hat{j} + 3\hat{k}$$

### Finding a Point on the Line

To define the line, we need a specific point on it. By setting $z = 0$, the equations of the planes become:

$$x - y = 2$$
$$2x + y = 2$$

Adding these equations yields $3x = 4$, so $x = \frac{4}{3}$. Substituting this into the first equation gives $y = -\frac{2}{3}$. Thus, a point on the line is $(\frac{4}{3}, -\frac{2}{3}, 0)$.

### Defining the Line and the Foot of the Perpendicular

The symmetric form of the line $L$ is:

$$\frac{x - \frac{4}{3}}{-1} = \frac{y + \frac{2}{3}}{5} = \frac{z - 0}{3} = \lambda$$

Any point $P$ on the line can be expressed in terms of $\lambda$ as:

$$P = \left( -\lambda + \frac{4}{3}, 5\lambda - \frac{2}{3}, 3\lambda ight)$$

### The Orthogonality Condition

Let $A = (1, 2, 0)$. The vector $\vec{AP}$ is given by:

$$\vec{AP} = \left( -\lambda + \frac{1}{3} ight)\hat{i} + \left( 5\lambda - \frac{8}{3} ight)\hat{j} + 3\lambda\hat{k}$$

Since $P$ is the foot of the perpendicular, $\vec{AP} \cdot \vec{d} = 0$:

$$-1\left( -\lambda + \frac{1}{3} ight) + 5\left( 5\lambda - \frac{8}{3} ight) + 3(3\lambda) = 0$$

Expanding and solving for $\lambda$:

$$\lambda - \frac{1}{3} + 25\lambda - \frac{40}{3} + 9\lambda = 0 \Rightarrow 35\lambda = \frac{41}{3} \Rightarrow \lambda = \frac{41}{105}$$

### Final Calculation

We require the value of $35(\alpha + \beta + \gamma)$, where $(\alpha, \beta, \gamma)$ are the coordinates of $P$. First, we sum the coordinates:

$$\alpha + \beta + \gamma = \left( -\lambda + \frac{4}{3} ight) + \left( 5\lambda - \frac{2}{3} ight) + 3\lambda = 7\lambda + \frac{2}{3}$$

Substituting $\lambda = \frac{41}{105}$:

$$\alpha + \beta + \gamma = 7\left( \frac{41}{105} ight) + \frac{2}{3} = \frac{41}{15} + \frac{10}{15} = \frac{51}{15} = \frac{17}{5}$$

The final result is:

$$35 	imes \left( \frac{17}{5} ight) = 7 	imes 17 = \mathbf{119}$$

Select Answer:

Visualized Solution

Intersection of Two Planes

Direction Vector of Line $L$

Calculating $d$

Finding a Point on Line $L$

Equation of Line $L$

The Perpendicular Condition

Vector $A P$

Applying the Dot Product

Solving for $λ$

Finding $α + β + γ$

Substituting $λ$

Final Answer

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Finding a Point on the Line

Defining the Line and the Foot of the Perpendicular

The Orthogonality Condition

Final Calculation

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Select Answer:

Visualized Solution

Intersection of Two Planes

Direction Vector of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

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

Finding a Point on Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L