Let `veca=4veci+3vecj and vecb=3veci-4vecj+5veck`. If `vecc` is a vector such that `vecc.(veca xx vecb)+25=0,vecc.(hati+hatj+hatk)=4`, and projection of `vecc` on `veca` is 1,then the projection of `vcec` on `vecb` equals

Options

A. `5/sqrt2`
B. `1/sqrt2`
C. `1/5`
D. `3/sqrt2`

Answer

To solve the given problem step by step, we will define the vectors, calculate necessary dot products, and find the required projections.

### Step 1: Define the vectors
Let:
\[ \vec{a} = 4\hat{i} + 3\hat{j} \]
\[ \vec{b} = 3\hat{i} - 4\hat{j} + 5\hat{k} \]
Let \(\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}\).

### Step 2: Use the condition \(\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4\)
This gives us:
\[ \vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = x + y + z = 4 \quad \text{(Equation 1)} \]

### Step 3: Calculate \(\vec{a} \times \vec{b}\)
Using the determinant method:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 0 \\ 3 & -4 & 5 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}(3 \cdot 5 - 0 \cdot (-4)) - \hat{j}(4 \cdot 5 - 0 \cdot 3) + \hat{k}(4 \cdot (-4) - 3 \cdot 3) \]
\[ = 15\hat{i} - 20\hat{j} - (16 + 9)\hat{k} \]
\[ = 15\hat{i} - 20\hat{j} - 25\hat{k} \]
Thus,
\[ \vec{a} \times \vec{b} = 15\hat{i} - 20\hat{j} - 25\hat{k} \]

### Step 4: Use the condition \(\vec{c} \cdot (\vec{a} \times \vec{b}) + 25 = 0\)
This gives us:
\[ \vec{c} \cdot (15\hat{i} - 20\hat{j} - 25\hat{k}) + 25 = 0 \]
\[ 15x - 20y - 25z + 25 = 0 \]
\[ 15x - 20y - 25z = -25 \quad \text{(Equation 2)} \]

### Step 5: Find the projection of \(\vec{c}\) on \(\vec{a}\)
The projection of \(\vec{c}\) on \(\vec{a}\) is given as 1:
\[ \text{Projection of } \vec{c} \text{ on } \vec{a} = \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} = 1 \]
Calculating \(|\vec{a}|\):
\[ |\vec{a}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5 \]
Thus,
\[ \vec{c} \cdot \vec{a} = 5 \]
Calculating \(\vec{c} \cdot \vec{a}\):
\[ \vec{c} \cdot \vec{a} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (4\hat{i} + 3\hat{j}) = 4x + 3y = 5 \quad \text{(Equation 3)} \]

### Step 6: Solve the equations
We have the following equations:
1. \(x + y + z = 4\) (Equation 1)
2. \(15x - 20y - 25z = -25\) (Equation 2)
3. \(4x + 3y = 5\) (Equation 3)

From Equation 3, express \(y\) in terms of \(x\):
\[ y = \frac{5 - 4x}{3} \]

Substituting \(y\) into Equation 1:
\[ x + \frac{5 - 4x}{3} + z = 4 \]
Multiplying through by 3 to eliminate the fraction:
\[ 3x + 5 - 4x + 3z = 12 \]
\[ -x + 3z = 7 \quad \text{(Equation 4)} \]

Now substitute \(y\) into Equation 2:
\[ 15x - 20\left(\frac{5 - 4x}{3}\right) - 25z = -25 \]
Multiplying through by 3:
\[ 45x - 100 + 80x - 75z = -75 \]
\[ 125x - 75z = 25 \quad \text{(Equation 5)} \]

### Step 7: Solve Equations 4 and 5
From Equation 4:
\[ 3z = x + 7 \implies z = \frac{x + 7}{3} \]
Substituting into Equation 5:
\[ 125x - 75\left(\frac{x + 7}{3}\right) = 25 \]
Multiplying through by 3:
\[ 375x - 75(x + 7) = 75 \]
\[ 375x - 75x - 525 = 75 \]
\[ 300x = 600 \implies x = 2 \]

### Step 8: Find \(y\) and \(z\)
Substituting \(x = 2\) back into Equation 3:
\[ 4(2) + 3y = 5 \implies 8 + 3y = 5 \implies 3y = -3 \implies y = -1 \]
Substituting \(x = 2\) and \(y = -1\) into Equation 1:
\[ 2 - 1 + z = 4 \implies z = 3 \]

### Step 9: Find \(\vec{c}\)
Thus,
\[ \vec{c} = 2\hat{i} - \hat{j} + 3\hat{k} \]

### Step 10: Find the projection of \(\vec{c}\) on \(\vec{b}\)
The projection of \(\vec{c}\) on \(\vec{b}\) is given by:
\[ \text{Projection of } \vec{c} \text{ on } \vec{b} = \frac{\vec{c} \cdot \vec{b}}{|\vec{b}|} \]
Calculating \(\vec{c} \cdot \vec{b}\):
\[ \vec{c} \cdot \vec{b} = (2\hat{i} - \hat{j} + 3\hat{k}) \cdot (3\hat{i} - 4\hat{j} + 5\hat{k}) = 2 \cdot 3 + (-1)(-4) + 3 \cdot 5 = 6 + 4 + 15 = 25 \]
Calculating \(|\vec{b}|\):
\[ |\vec{b}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \]
Thus,
\[ \text{Projection of } \vec{c} \text{ on } \vec{b} = \frac{25}{5\sqrt{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2} \]

### Final Answer
The projection of \(\vec{c}\) on \(\vec{b}\) equals \(\frac{5\sqrt{2}}{2}\).

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