What is the number of vectors of length 5 units, perpendicular to the vectors `vec(a) = (1.1.0) and vec(b)= (0.1.1)` ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem of finding the number of vectors of length 5 units that are perpendicular to the vectors \(\vec{a} = (1, 1, 0)\) and \(\vec{b} = (0, 1, 1)\), we can follow these steps:

### Step 1: Define the vector
Let the vector we are looking for be represented as:
\[ \vec{v} = (a, b, c) \]

### Step 2: Set up the perpendicularity conditions
For \(\vec{v}\) to be perpendicular to \(\vec{a}\), the dot product must equal zero:
\[ \vec{v} \cdot \vec{a} = a \cdot 1 + b \cdot 1 + c \cdot 0 = a + b = 0 \]
This gives us our first equation:
\[ b = -a \quad \text{(1)} \]

Next, for \(\vec{v}\) to be perpendicular to \(\vec{b}\), we set up the second dot product:
\[ \vec{v} \cdot \vec{b} = a \cdot 0 + b \cdot 1 + c \cdot 1 = b + c = 0 \]
This gives us our second equation:
\[ c = -b \quad \text{(2)} \]

### Step 3: Substitute the equations
From equation (1), we substitute \(b\) into equation (2):
\[ c = -(-a) = a \quad \text{(3)} \]

### Step 4: Express everything in terms of \(a\)
Now we can express \(b\) and \(c\) in terms of \(a\):
\[ b = -a \quad \text{and} \quad c = a \]
Thus, we can rewrite the vector \(\vec{v}\) as:
\[ \vec{v} = (a, -a, a) \]

### Step 5: Set up the length condition
The length of the vector \(\vec{v}\) must be 5 units:
\[ \sqrt{a^2 + (-a)^2 + a^2} = 5 \]
This simplifies to:
\[ \sqrt{3a^2} = 5 \]
Squaring both sides gives:
\[ 3a^2 = 25 \]
Thus,
\[ a^2 = \frac{25}{3} \]
Taking the square root, we find:
\[ a = \pm \frac{5}{\sqrt{3}} \]

### Step 6: Find the corresponding vectors
Using the values of \(a\):
1. If \(a = \frac{5}{\sqrt{3}}\):
\[ \vec{v_1} = \left(\frac{5}{\sqrt{3}}, -\frac{5}{\sqrt{3}}, \frac{5}{\sqrt{3}}\right) \]
2. If \(a = -\frac{5}{\sqrt{3}}\):
\[ \vec{v_2} = \left(-\frac{5}{\sqrt{3}}, \frac{5}{\sqrt{3}}, -\frac{5}{\sqrt{3}}\right) \]

### Step 7: Conclusion
Thus, we have found two vectors of length 5 units that are perpendicular to both \(\vec{a}\) and \(\vec{b}\).

### Final Answer
The number of vectors of length 5 units that are perpendicular to \(\vec{a}\) and \(\vec{b}\) is **2**.

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