Let P be a plane passing through the points (1, 0, 1), (1, -2, 1) and (0, 1, -2). Let a vector a = α

Q: Wait, if $\vec{a}$ is parallel to the plane, why is its dot product with the normal vector zero? Shouldn't it be parallel to the normal?

That is a classic confusion! Remember, the normal vector $\vec{n}$ is perpendicular to every single line and vector lying on the plane. If $\vec{a}$ is parallel to the plane, it lies on the plane (or is parallel to it), so it must be perpendicular to the normal vector. Therefore, $\vec{a} \cdot \vec{n} = 0$ is the correct condition.

Q: Can I pick any two vectors from the three points to find the normal vector?

Absolutely! Any two non-collinear vectors formed by the points $A, B,$ and $C$ will span the same plane. Whether you use $\vec{AB}$ and $\vec{AC}$, or $\vec{BA}$ and $\vec{BC}$, the resulting cross product will always be parallel to the same normal vector $\vec{n}$. The direction ratios will be the same, which is all that matters for the perpendicularity condition.

Q: Why do we need to solve for $\alpha, \beta,$ and $\gamma$ individually? Can't we just find the expression $(\alpha - \beta + \gamma)^2$ directly?

While the expression is linear in $\alpha, \beta,$ and $\gamma$, we have three independent conditions that uniquely determine the vector $\vec{a}$. Without solving for the individual components, we wouldn't have enough information to evaluate the final squared expression. Think of it as solving a puzzle: you need all the pieces to see the full picture!

Question

Let $P$ be a plane passing through the points $(1, 0, 1), (1, -2, 1)$ and $(0, 1, -2)$. Let a vector $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ be such that $\vec{a}$ is parallel to the plane $P$, perpendicular to $(\hat{i} + 2\hat{j} + 3\hat{k})$ and $\vec{a} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 2$, then $(\alpha - \beta + \gamma)^2$ equals ___

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To define the plane $P$, we identify two vectors lying within it using the given points $A(1, 0, 1)$, $B(1, -2, 1)$, and $C(0, 1, -2)$.

We calculate these vectors as:
$$\vec{AB} = B - A = (0, -2, 0)$$
$$\vec{AC} = C - A = (-1, 1, -3)$$

The normal vector $\vec{n}$ to the plane is found via the cross product $\vec{n} = \vec{AB} 	imes \vec{AC}$:
$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -2 & 0 \ -1 & 1 & -3 \end{vmatrix} = 6\hat{i} + 0\hat{j} - 2\hat{k}$$

For algebraic simplicity, we scale this vector to the direction ratio $(3, 0, -1)$.

### The Mystery Vector and Constraints

We define the mystery vector as $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$.

Since $\vec{a}$ is parallel to the plane $P$, it must be perpendicular to the normal vector $\vec{n}$. This implies $\vec{a} \cdot \vec{n} = 0$:
$$3\alpha + 0\beta - \gamma = 0 \implies \gamma = 3\alpha$$

The second condition states that $\vec{a}$ is perpendicular to $\vec{v} = \hat{i} + 2\hat{j} + 3\hat{k}$. Applying the dot product $\vec{a} \cdot \vec{v} = 0$:
$$\alpha + 2\beta + 3\gamma = 0$$

### The Algebraic Symphony

We substitute $\gamma = 3\alpha$ into the second constraint:
$$\alpha + 2\beta + 3(3\alpha) = 0$$
$$10\alpha + 2\beta = 0 \implies \beta = -5\alpha$$

We now utilize the final condition: $\vec{a} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 2$. Substituting our expressions for $\beta$ and $\gamma$ in terms of $\alpha$:
$$\alpha(1) + (-5\alpha)(1) + (3\alpha)(2) = 2$$
$$\alpha - 5\alpha + 6\alpha = 2$$
$$2\alpha = 2 \implies \alpha = 1$$

With $\alpha = 1$, we find the remaining components:
$$\beta = -5(1) = -5$$
$$\gamma = 3(1) = 3$$

### The Grand Finale

The vector is $\vec{a} = (1, -5, 3)$. We are tasked with calculating $(\alpha - \beta + \gamma)^2$:
$$(\alpha - \beta + \gamma)^2 = (1 - (-5) + 3)^2$$
$$(1 + 5 + 3)^2 = 9^2$$

The final result is **81**.

Enter Numerical Value:

Visualized Solution

Identifying Points on Plane $P$

Finding Vectors in Plane $P$

Calculating Normal Vector $n$

Condition 1: $a$ is Parallel to Plane $P$

Applying Condition 1

Condition 2: $a ⊥ (\hat{i} + 2 \hat{j} + 3 \hat{k})$

Substituting $γ$ into Condition 2

Solving for $β$

Condition 3: Final Dot Product

Substituting $β$ and $γ$

Finding $α, β, γ$

Calculating Final Expression

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Mystery Vector and Constraints

The Algebraic Symphony

The Grand Finale

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Enter Numerical Value:

Visualized Solution

Identifying Points on Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Finding Vectors in Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

Condition 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a is Parallel to Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Applying Condition 1

Condition 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a⊥(i^+2j^​+3k^)

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ into Condition 2

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