If `vec(a)=(4hat(i)+ 5 hat(j) - hat(k)),vec(b)=(hat(i)-4 hat(j)+ 5 hat(k))`, and`vec(c)=(3 hat(i)+ hat(j)-hat(k))` find a vector `vec(d)` which is perpendicular to both `vec(a)` and `vec(b)` and for which `vec(c )*vec(d)=21.`

Answer

To solve the problem, we need to find a vector \( \vec{d} \) that is perpendicular to both vectors \( \vec{a} \) and \( \vec{b} \), and also satisfies the condition \( \vec{c} \cdot \vec{d} = 21 \).

### Step 1: Find the Cross Product \( \vec{a} \times \vec{b} \)

The first step is to calculate the cross product \( \vec{a} \times \vec{b} \), which will give us a vector that is perpendicular to both \( \vec{a} \) and \( \vec{b} \).

Given:
\[ \vec{a} = 4\hat{i} + 5\hat{j} - \hat{k} \]
\[ \vec{b} = \hat{i} - 4\hat{j} + 5\hat{k} \]

We can use the determinant method to find the cross product:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & -1 \\ 1 & -4 & 5 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 5 & -1 \\ -4 & 5 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & -1 \\ 1 & 5 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 5 \\ 1 & -4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 5 & -1 \\ -4 & 5 \end{vmatrix} = (5)(5) - (-1)(-4) = 25 - 4 = 21 \)
2. \( \begin{vmatrix} 4 & -1 \\ 1 & 5 \end{vmatrix} = (4)(5) - (-1)(1) = 20 + 1 = 21 \)
3. \( \begin{vmatrix} 4 & 5 \\ 1 & -4 \end{vmatrix} = (4)(-4) - (5)(1) = -16 - 5 = -21 \)

Putting it all together:
\[ \vec{a} \times \vec{b} = 21\hat{i} - 21\hat{j} - 21\hat{k} \]

### Step 2: Express \( \vec{d} \)

Since \( \vec{d} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \), we can express \( \vec{d} \) as:
\[ \vec{d} = \lambda (21\hat{i} - 21\hat{j} - 21\hat{k}) = 21\lambda (\hat{i} - \hat{j} - \hat{k}) \]

### Step 3: Use the Dot Product Condition

Now, we need to satisfy the condition \( \vec{c} \cdot \vec{d} = 21 \).

Given:
\[ \vec{c} = 3\hat{i} + \hat{j} - \hat{k} \]

Calculating the dot product:
\[ \vec{c} \cdot \vec{d} = (3\hat{i} + \hat{j} - \hat{k}) \cdot (21\lambda (\hat{i} - \hat{j} - \hat{k})) \]
\[ = 21\lambda (3 \cdot 1 + 1 \cdot (-1) + (-1) \cdot (-1)) = 21\lambda (3 - 1 + 1) = 21\lambda \cdot 3 = 63\lambda \]

Setting this equal to 21:
\[ 63\lambda = 21 \]
\[ \lambda = \frac{21}{63} = \frac{1}{3} \]

### Step 4: Find \( \vec{d} \)

Substituting \( \lambda \) back into the expression for \( \vec{d} \):
\[ \vec{d} = 21 \cdot \frac{1}{3} (\hat{i} - \hat{j} - \hat{k}) = 7(\hat{i} - \hat{j} - \hat{k}) \]

### Final Answer

Thus, the vector \( \vec{d} \) is:
\[ \vec{d} = 7\hat{i} - 7\hat{j} - 7\hat{k} \]

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