Given two vectors
`veca=2hati-3hatj+6hatk, vecb=-2hati+2hatj-hatk` and
`lambda=("the projection of "veca on vecb)/("the projection of " vecb on veca)`, then the value of `lambda` is

Options

A. `(3)/(7)`
B. 7
C. 3
D. `(7)/(3)`

Answer

To find the value of \( \lambda \), which is defined as the ratio of the projection of vector \( \vec{a} \) on vector \( \vec{b} \) to the projection of vector \( \vec{b} \) on vector \( \vec{a} \), we can follow these steps:

### Step 1: Write down the vectors
Given:
\[ \vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k} \]
\[ \vec{b} = -2\hat{i} + 2\hat{j} - \hat{k} \]

### Step 2: Calculate the dot product \( \vec{a} \cdot \vec{b} \)
The dot product is calculated as follows:
\[ \vec{a} \cdot \vec{b} = (2)(-2) + (-3)(2) + (6)(-1) \]
\[ = -4 - 6 - 6 = -16 \]

### Step 3: Calculate the magnitudes of \( \vec{a} \) and \( \vec{b} \)
**Magnitude of \( \vec{a} \)**:
\[ |\vec{a}| = \sqrt{(2)^2 + (-3)^2 + (6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

**Magnitude of \( \vec{b} \)**:
\[ |\vec{b}| = \sqrt{(-2)^2 + (2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 4: Calculate the projections
**Projection of \( \vec{a} \) on \( \vec{b} \)**:
\[ \text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} = \frac{-16}{3^2} \vec{b} = \frac{-16}{9} \vec{b} \]

**Projection of \( \vec{b} \) on \( \vec{a} \)**:
\[ \text{proj}_{\vec{a}} \vec{b} = \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \vec{a} = \frac{-16}{7^2} \vec{a} = \frac{-16}{49} \vec{a} \]

### Step 5: Calculate \( \lambda \)
Now, we can find \( \lambda \):
\[ \lambda = \frac{\text{proj}_{\vec{b}} \vec{a}}{\text{proj}_{\vec{a}} \vec{b}} = \frac{\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}}{\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}} = \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{a}} \cdot \frac{|\vec{a}|}{|\vec{b}|} \]
Since \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \), they cancel out:
\[ \lambda = \frac{|\vec{a}|}{|\vec{b}|} = \frac{7}{3} \]

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \lambda = \frac{7}{3} \]

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