A vector of magnitude 9 perpendicular to both the vectors `a = 4i - j+k` and `b =-2i+j-2k` is `-3i+6j+6k` . True or False

Answer

To determine if the vector \(\mathbf{c} = -3\mathbf{i} + 6\mathbf{j} + 6\mathbf{k}\) is perpendicular to both vectors \(\mathbf{a} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}\) and \(\mathbf{b} = -2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\), we will follow these steps:

### Step 1: Calculate the dot product of \(\mathbf{a}\) and \(\mathbf{c}\)

The dot product of two vectors \(\mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k}\) and \(\mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k}\) is given by:

\[ \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \]

For \(\mathbf{a} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}\) and \(\mathbf{c} = -3\mathbf{i} + 6\mathbf{j} + 6\mathbf{k}\):

\[ \mathbf{a} \cdot \mathbf{c} = (4)(-3) + (-1)(6) + (1)(6) \]

Calculating this gives:

\[ \mathbf{a} \cdot \mathbf{c} = -12 - 6 + 6 = -12 \]

### Step 2: Check if the dot product is zero

Since \(\mathbf{a} \cdot \mathbf{c} = -12\), it is not equal to zero. Therefore, \(\mathbf{c}\) is not perpendicular to \(\mathbf{a}\).

### Step 3: Calculate the dot product of \(\mathbf{b}\) and \(\mathbf{c}\)

Now, we will calculate the dot product of \(\mathbf{b} = -2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\) and \(\mathbf{c} = -3\mathbf{i} + 6\mathbf{j} + 6\mathbf{k}\):

\[ \mathbf{b} \cdot \mathbf{c} = (-2)(-3) + (1)(6) + (-2)(6) \]

Calculating this gives:

\[ \mathbf{b} \cdot \mathbf{c} = 6 + 6 - 12 = 0 \]

### Step 4: Check if the dot product is zero

Since \(\mathbf{b} \cdot \mathbf{c} = 0\), this means that \(\mathbf{c}\) is perpendicular to \(\mathbf{b}\).

### Step 5: Conclusion

Since \(\mathbf{c}\) is not perpendicular to \(\mathbf{a}\) but is perpendicular to \(\mathbf{b}\), we conclude that the statement "A vector of magnitude 9 perpendicular to both the vectors \(\mathbf{a}\) and \(\mathbf{b}\) is \(-3\mathbf{i} + 6\mathbf{j} + 6\mathbf{k}\)" is **False**.