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)
- What is the remainder when 2023²⁰²⁴ + 2025²⁰²⁴ is divided by 2024?
In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If she/he attempts all 60 questions and secures 130 marks, the number of questions she/he attempts wrongly, are?
Match List-I with List-II
List-1 List-II (A) If $\begin{bmatrix}\lambda-1 & 0 \\ 0 & \lambda-1 \end{bmatrix} $, then $\lambda$ is (I) 0 (B) If A=$ \begin{bmatrix}1 & 2 \\2 & 4 \end{bmatrix} $, then $\Delta$ is (II) 1 (C) If A = $ \begin{bmatrix}1 & 0 \\0 & \frac{1}{2} \end{bmatrix} $, then $|A^{-1}|$ is (III) -2 (D) If $ \begin{bmatrix}a+1 & 1 \\1 & 2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\1 & 2 \end{bmatrix} $, then a is (IV) 2 Choose the correct answer from the options given below:
- If (x - 1) is a factor of $2x^2 - 5x + k = 0$, then the value of k is:
- If $x = (2+\sqrt{3})^{\frac{1}{3}} + (2+\sqrt{3})^{-\frac{1}{3}}$ and $x^3-3x + k = 0$, then the value of k is:
