Vector Normality to Plane Equation
The standard vector equation for a plane is represented as $N \cdot X = d$. In this equation:
- $N$ represents the normal vector, which is perpendicular to the plane.
- $X$ is the position vector of any point $(x, y, z)$ lying on the plane.
- $d$ is a constant scalar value.
The given plane equation is $W^T \cdot X = 1$. By comparing this to the general form $N \cdot X = d$, we can identify that the vector $W^T$ corresponds to the normal vector $N$. The constant $d$ is equal to $1$.
We are provided with the vector $W^T = [1, 2, 3]$. Consequently, the normal vector to the plane defined by $W^T \cdot X = 1$ is this vector $W^T$ itself, which is $[1, 2, 3]^T$.
Identifying the Matching Vector Option
The question asks us to identify the vector that is normal to the plane. We have established that the normal vector is $W^T = [1, 2, 3]^T$. We now need to find which of the given options matches this vector.
| Option Number | Vector Given | Is it Equal to $[1, 2, 3]^T$? |
|---|---|---|
| 1 | $[-2, -2, 2]^T$ |
No |
| 2 | $[3, 0, -1]^T$ |
No |
| 3 | $[3, 2, 1]^T$ |
No |
| 4 | $[1, 2, 3]^T$ |
Yes |
| 5 | (No vector specified) | N/A |
The comparison shows that Option 4, represented by the vector $[1, 2, 3]^T$, is the one that exactly matches the calculated normal vector $W^T$.
Important Questions from Algebra (Notes)
- If $(y-12) = 4\sqrt{5}$, then find the value of $\sqrt{y-3} - \frac{1}{\sqrt{y-3}}$.
- In the expansion of (x + 9)(x - 6)(x + 5), what is the coefficient of x?
- The roots of the equation $ax^3-24x^2+188x-480=0$ are three consecutive even natural numbers. The value of a is _____.
- A square matrix having all the elements above the leading diagonal equal to zero is known as:
- The difference between two numbers is 16. If one-third of the smaller number is greater than one-seventh of the larger number by 4, then what is the larger number?
