The distance of the point having position vector -i + 2j + 6k from the straight line passing through

Q: Wait, why did the dot product come out negative? Did I mess up the calculation?

Not at all! The dot product is a scalar that tells us about the angle between vectors. A negative value just means the angle is obtuse. Since we are dealing with distance, we take the absolute value, so the sign doesn't affect the final length.

Q: Why can't we just use the cross product formula for distance?

You absolutely can! The formula $d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$ is a fantastic shortcut. However, the projection method we used is often safer because it builds intuition about the geometry of the triangle. Both methods will lead you to the same beautiful answer of 7.

Q: Is there a way to check if my answer is correct?

Yes! You can find the coordinates of the foot of the perpendicular $D$ by adding the projection vector to $A$. If the vector $\vec{PD}$ is perpendicular to $\vec{b}$ (i.e., their dot product is zero), you know you've nailed it.

Question

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 :

Options:

(A) 7

(B) 4\sqrt{3}

(C) 2\sqrt{13}

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the 3D coordinate system. Today, we are mapping a landscape where a point $P$ floats in 3D space, and a straight line passes through point $A$ in the direction of vector $\vec{b}$.

Our mission is to find the shortest distance from $P$ to this line. Geometrically, this is equivalent to dropping a plumb line from $P$ to the line such that it hits at a right angle, forming the distance $PD$.

### The Vector Toolkit

We define point $P$ at $(-1, 2, 6)$ and point $A$ at $(2, 3, -4)$. To bridge the gap between them, we construct the vector $\vec{AP}$:

$$\vec{AP} = \vec{P} - \vec{A} = (-1 - 2)\hat{i} + (2 - 3)\hat{j} + (6 - (-4))\hat{k} = -3\hat{i} - \hat{j} + 10\hat{k}$$

The magnitude of this vector, which serves as the hypotenuse of our geometric triangle, is calculated as follows:

$$|\vec{AP}| = \sqrt{(-3)^2 + (-1)^2 + 10^2} = \sqrt{9 + 1 + 100} = \sqrt{110}$$

### The Shadow of the Vector

To find the base of our triangle, we project $\vec{AP}$ onto the direction vector $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$. First, we determine the magnitude of $\vec{b}$:

$$|\vec{b}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$$

Next, we compute the dot product $\vec{AP} \cdot \vec{b}$:

$$\vec{AP} \cdot \vec{b} = (-3)(6) + (-1)(3) + (10)(-4) = -18 - 3 - 40 = -61$$

The length of the projection, $AD$, is the absolute value of this dot product divided by the magnitude of $\vec{b}$:

$$AD = \frac{|-61|}{\sqrt{61}} = \sqrt{61}$$

### The Pythagorean Finale

We now consider the right-angled triangle $APD$, where $AP$ is the hypotenuse, $AD$ is the base, and $PD$ is the perpendicular height. By the Pythagorean theorem, we have:

$$PD^2 + AD^2 = |\vec{AP}|^2$$

Rearranging to solve for our target distance $PD$:

$$PD^2 = |\vec{AP}|^2 - AD^2$$

Substituting our calculated values:

$$PD^2 = (\sqrt{110})^2 - (\sqrt{61})^2 = 110 - 61 = 49$$

Taking the square root, we find the final result:

**$PD = 7$**

Animated Solution for Mathematics - Three Dimensional Geometry: 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 :

Select Answer:

Visualized Solution

Visualizing the Geometry

The Strategy: Projection Method

Constructing Vector $A P$

Calculating Magnitude of $A P$

Magnitude of Direction Vector $b$

Calculating the Dot Product $A P \cdot b$

Finding Projection Length $A D$

Applying Pythagoras Theorem

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Vector Toolkit

The Shadow of the Vector

The Pythagorean Finale

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Select Answer:

Visualized Solution

Visualizing the Geometry

The Strategy: Projection Method

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

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

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

Calculating the Dot Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AP⋅b

Finding Projection Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AD