Show that If `vec(a)=x hat(i)+2hat(j)-z hat(k)` and `vec(b)=3hat(i)-y hat(j)+hat(k)` are two equal vectors, then `x+y+z=0`.

Answer

To show that if \(\vec{a} = x \hat{i} + 2 \hat{j} - z \hat{k}\) and \(\vec{b} = 3 \hat{i} - y \hat{j} + \hat{k}\) are two equal vectors, then \(x + y + z = 0\), we will follow these steps:

### Step 1: Set the vectors equal to each other
Since \(\vec{a}\) and \(\vec{b}\) are equal, we can write:
\[ x \hat{i} + 2 \hat{j} - z \hat{k} = 3 \hat{i} - y \hat{j} + \hat{k} \]

### Step 2: Equate the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\)
From the equality of the vectors, we can equate the coefficients of each unit vector:
1. Coefficient of \(\hat{i}\):
\[ x = 3 \tag{1} \]
2. Coefficient of \(\hat{j}\):
\[ 2 = -y \tag{2} \]
3. Coefficient of \(\hat{k}\):
\[ -z = 1 \tag{3} \]

### Step 3: Solve for \(y\) and \(z\)
From equation (2):
\[ y = -2 \]

From equation (3):
\[ z = -1 \]

### Step 4: Substitute the values of \(x\), \(y\), and \(z\) into \(x + y + z\)
Now we substitute the values we found:
\[ x + y + z = 3 + (-2) + (-1) \]

### Step 5: Simplify the expression
Calculating the above expression:
\[ x + y + z = 3 - 2 - 1 = 0 \]

### Conclusion
Thus, we have shown that \(x + y + z = 0\).

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