Let A, B, C be three points whose position vectors respectively are: a = i + 4j + 3k, b = 2i + αj +

Q: Wait, why do we reject the value $\alpha = 1$?

Great catch! The problem explicitly states that $A, B,$ and $C$ form a triangle. If $\alpha = 1$, the displacement vectors $\vec{AB}$ and $\vec{AC}$ become proportional, meaning the points lie on a single straight line. Since a triangle cannot exist with collinear vertices, we must exclude $\alpha = 1$ from our set of possible values.

Q: Why did we choose $\alpha = 2$?

The problem asks for the *smallest positive integer* $\alpha$ such that the points are non-collinear. We found that the points are collinear only when $\alpha = 1$. The set of positive integers is $\{1, 2, 3, \dots\}$. Since $\alpha$ cannot be $1$, the next smallest integer in the set is $2$. Thus, $\alpha = 2$ is the correct choice.

Q: Is there a faster way to find the length of the median?

While you could use the vector formula for the median $\vec{AM} = \frac{1}{2}(\vec{AB} + \vec{AC})$, calculating the coordinates of the midpoint $M$ and then using the distance formula is often less prone to algebraic errors in 3D geometry. Both methods will yield the same result, but the coordinate approach is very intuitive.

Question

Let $A, B, C$ be three points whose position vectors respectively are: $\vec{a} = \hat{i} + 4\hat{j} + 3\hat{k}$, $\vec{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}, \alpha \in \mathbb{R}$, $\vec{c} = 3\hat{i} - \2\hat{j} + 5\hat{k}$. If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear, then the length of the median, in $	riangle ABC$, through $A$ is:

Options:

(A) $\frac{\sqrt{82}}{2}$

(B) $\frac{\sqrt{62}}{2}$

(C) $\frac{\sqrt{69}}{2}$

(D) $\frac{\sqrt{66}}{2}$

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: A Journey Through Space

Imagine you are standing in a three-dimensional coordinate system. You have three points, $A, B,$ and $C$, floating in space.

Their positions are defined by vectors $\vec{a} = \hat{i} + 4\hat{j} + 3\hat{k}$, $\vec{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$, and $\vec{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$.

We are tasked with finding the length of the median through $A$, given that the position of point $B$ depends on an unknown parameter $\alpha$. Our journey begins by unlocking the secret of this parameter.

### Phase 1: The Collinearity Trap

To form a triangle, our three points must be non-collinear. If they were collinear, they would simply form a line segment, not a triangle.

We test for this by examining the displacement vectors $\vec{AB}$ and $\vec{AC}$. If these two vectors are parallel, the points lie on the same line.

Let us calculate them:

$$\vec{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (\alpha-4)\hat{j} + (4-3)\hat{k} = \hat{i} + (\alpha-4)\hat{j} + \hat{k}$$

$$\vec{AC} = \vec{c} - \vec{a} = (3-1)\hat{i} + (-2-4)\hat{j} + (5-3)\hat{k} = 2\hat{i} - 6\hat{j} + 2\hat{k}$$

For these vectors to be parallel, their direction ratios must be proportional. This gives us the condition:

$$\frac{1}{2} = \frac{\alpha-4}{-6} = \frac{1}{2}$$

Solving $\frac{\alpha-4}{-6} = \frac{1}{2}$ leads us to $\alpha - 4 = -3$, which means $\alpha = 1$. This is the critical value where the triangle collapses into a line.

Since the problem demands a triangle, we know $\alpha 
eq 1$. We are looking for the smallest positive integer $\alpha$. Since $1$ is excluded, the smallest positive integer is $\alpha = 2$.

With this, our point $B$ is locked at $(2, 2, 4)$.

### Phase 2: The Geometry of the Median

A median is a bridge between a vertex and the heart of the opposite side. Specifically, the median through $A$ connects vertex $A$ to the midpoint $M$ of side $BC$.

To find $M$, we simply average the coordinates of $B(2, 2, 4)$ and $C(3, -2, 5)$:

$$M = \left( \frac{2+3}{2}, \frac{2+(-2)}{2}, \frac{4+5}{2} ight) = \left( \frac{5}{2}, 0, \frac{9}{2} ight)$$

Now, we have our two anchor points: $A(1, 4, 3)$ and $M(\frac{5}{2}, 0, \frac{9}{2})$. The final step is to calculate the distance $AM$ using the 3D distance formula:

$$AM = \sqrt{\left(\frac{5}{2}-1ight)^2 + (0-4)^2 + \left(\frac{9}{2}-3ight)^2}$$

This simplifies to:

$$AM = \sqrt{\left(\frac{3}{2}ight)^2 + (-4)^2 + \left(\frac{3}{2}ight)^2} = \sqrt{\frac{9}{4} + 16 + \frac{9}{4}}$$

Combining the fractions, we get $\frac{18}{4} = \frac{9}{2}$. Thus, $AM = \sqrt{\frac{9}{2} + 16} = \sqrt{\frac{9+32}{2}} = \sqrt{\frac{41}{2}}$.

To match our standard form, we rationalize the denominator by multiplying by $\frac{\sqrt{2}}{\sqrt{2}}$:

$$AM = \sqrt{\frac{82}{4}} = \frac{\sqrt{82}}{2}$$

### Conclusion

Through the logic of vectors and the precision of the distance formula, we have unraveled the mystery.

The length of the median is **$\frac{\sqrt{82}}{2}$**.

It is a reminder that in geometry, every variable has a purpose, and every constraint is a clue waiting to be solved. Keep practicing, and keep visualizing the space around you!

Select Answer:

Visualized Solution

Visualizing the Points $A, B, C$

Condition for Non-Collinearity

Calculating Vectors $A B$ and $A C$

Finding the Collinear Case

Determining the Smallest Integer $α$

The Median Through $A$

Finding the Midpoint $M$

Setting up the Distance Formula for $A M$

Calculating the Length $A M$

Final Simplification

The Sigma Insight: Components of a Vector

The Geometry of Vectors

Phase 1

Phase 2

Conclusion

Similar Questions

Let $α, β, γ$ be distinct real numbers. The points with position vectors $α \hat{i} + β \hat{j} + γ \hat{k}$ , $β \hat{i} + γ \hat{j} + α \hat{k}$ , $γ \hat{i} + α \hat{j} + β \hat{k}$

A vector $a = α \hat{i} + 2 \hat{j} + β \hat{k} (α, β \in R)$ lies in the plane of the vectors, $b = \hat{i} + \hat{j}$ and $c = \hat{i} - \hat{j} + 4 \hat{k}$ . If $a$ bisects the angle between $b$ and $c$ , then

If the components of $a = α \hat{i} + β \hat{j} + γ \hat{k}$ along and perpendicular to $b = 3 \hat{i} + \hat{j} - \hat{k}$ respectively, are $\frac{16}{11} (3 \hat{i} + \hat{j} - \hat{k})$ and $\frac{1}{11} (- 4 \hat{i} - 5 \hat{j} - 17 \hat{k})$ , then $α^{2} + β^{2} + γ^{2}$ is equal to :

Match the following:

List-I

List-II

If $b$ and $c$ are any two non-collinear unit vectors and $a$ is any vector, then $(a \cdot b) b + (a \cdot c) c + \frac{a \cdot ( b \times c )}{∣ b \times c ∣ ^{2}} (b \times c) = .........$

The points with position vectors $60 \hat{i} + 3 \hat{j}, 40 \hat{i} - 8 \hat{j}, a \hat{i} - 52 \hat{j}$ are collinear if

For any vector $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ , with $10∣ a_{i} ∣ < 1, i = 1, 2, 3$ , consider the following statements: (A) $max {∣ a_{1} ∣, ∣ a_{2} ∣, ∣ a_{3} ∣} \leq ∣ a ∣$ , (B) $∣ a ∣ \leq 3 max {∣ a_{1} ∣, ∣ a_{2} ∣, ∣ a_{3} ∣}$

Let $a = 2 \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2 \hat{j} - \hat{k}$ and a unit vector $c$ be coplanar. If $c$ is perpendicular to $a$ , then $c =$

Let $α = 4 \hat{i} + 3 \hat{j} + 5 \hat{k}$ and $β = \hat{i} + 2 \hat{j} - 4 \hat{k}$ . Let $β_{1}$ be parallel to $α$ and $β_{2}$ be perpendicular to $α$ . If $β = β_{1} + β_{2}$ , then the value of $5 β_{2} \cdot (\hat{i} + \hat{j} + \hat{k})$ is

Select Answer:

Visualized Solution

Visualizing the Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A,B,C

Condition for Non-Collinearity

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

Finding the Collinear Case

Determining the Smallest Integer .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

The Median Through .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding the Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Setting up the Distance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AM

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