If `veca=4hatj+5hatj-hatk,vecb=hati-4hatj+4hatk` and `vecc = 3hati+hatj+hatk` Find a vector `vecd` which is perpendicualr to both `veca` and `vecb` and `vecd.vecc` = 21 .

Answer

To find the vector \(\vec{d}\) that is perpendicular to both \(\vec{a}\) and \(\vec{b}\), and satisfies \(\vec{d} \cdot \vec{c} = 21\), we can follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = 4\hat{j} + 5\hat{j} - \hat{k} = 0\hat{i} + 9\hat{j} - \hat{k} \]
\[ \vec{b} = \hat{i} - 4\hat{j} + 4\hat{k} \]
\[ \vec{c} = 3\hat{i} + \hat{j} + \hat{k} \]

### Step 2: Assume the form of \(\vec{d}\)
Let:
\[ \vec{d} = x\hat{i} + y\hat{j} + z\hat{k} \]

### Step 3: Set up the equations
1. Since \(\vec{d}\) is perpendicular to \(\vec{a}\):
\[ \vec{d} \cdot \vec{a} = 0 \implies 0\cdot x + 9y - z = 0 \implies 9y - z = 0 \quad \text{(Equation 1)} \]

2. Since \(\vec{d}\) is perpendicular to \(\vec{b}\):
\[ \vec{d} \cdot \vec{b} = 0 \implies x - 4y + 4z = 0 \quad \text{(Equation 2)} \]

3. From the condition \(\vec{d} \cdot \vec{c} = 21\):
\[ \vec{d} \cdot \vec{c} = 21 \implies 3x + y + z = 21 \quad \text{(Equation 3)} \]

### Step 4: Solve the equations
From Equation 1:
\[ z = 9y \]

Substituting \(z\) in Equation 2:
\[ x - 4y + 4(9y) = 0 \implies x - 4y + 36y = 0 \implies x + 32y = 0 \implies x = -32y \quad \text{(Equation 4)} \]

Now substitute \(x\) and \(z\) in Equation 3:
\[ 3(-32y) + y + 9y = 21 \]
\[ -96y + 10y = 21 \implies -86y = 21 \implies y = -\frac{21}{86} \]

Using \(y\) to find \(x\) and \(z\):
\[ x = -32\left(-\frac{21}{86}\right) = \frac{672}{86} \]
\[ z = 9\left(-\frac{21}{86}\right) = -\frac{189}{86} \]

### Step 5: Write the vector \(\vec{d}\)
Thus, we have:
\[ \vec{d} = \frac{672}{86}\hat{i} - \frac{21}{86}\hat{j} - \frac{189}{86}\hat{k} \]

### Final Answer
\[ \vec{d} = \frac{672}{86}\hat{i} - \frac{21}{86}\hat{j} - \frac{189}{86}\hat{k} \]