If i-3j+5k bisects angle between a and -i+2j+2k; where a is unit vector than a=?

Let **a** be a unit vector that satisfies the given condition.

### Step 1: Understanding the Condition
The given vector **v** = **i - 3j + 5k** bisects the angle between **a** and **b**, where
**b** = **-i + 2j + 2k**.

Since **v** is the angle bisector, it satisfies the vector angle bisector condition:

\[
\frac{a}{|a|} + \frac{b}{|b|} = \lambda v
\]

where **λ** is some scalar.

Since **a** is a unit vector, **|a| = 1**, so the equation simplifies to:

\[
a + \frac{b}{|b|} = \lambda v
\]

### Step 2: Finding |b|
We calculate the magnitude of **b**:

\[
|b| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3
\]

Thus, the unit vector in the direction of **b** is:

\[
\frac{b}{|b|} = \frac{-i + 2j + 2k}{3} = -\frac{1}{3}i + \frac{2}{3}j + \frac{2}{3}k
\]

### Step 3: Expressing a in Terms of λ
Substituting the unit vector of **b** into the bisector equation:

\[
a + \left(-\frac{1}{3}i + \frac{2}{3}j + \frac{2}{3}k\right) = \lambda (i - 3j + 5k)
\]

Rearrange:

\[
a = \lambda (i - 3j + 5k) + \frac{1}{3}i - \frac{2}{3}j - \frac{2}{3}k
\]

### Step 4: Applying the Unit Vector Condition
Since **a** is a unit vector, its magnitude must be 1:

\[
\left| \lambda (i - 3j + 5k) + \frac{1}{3}i - \frac{2}{3}j - \frac{2}{3}k \right| = 1
\]

Expanding component-wise:

\[
a_x = \lambda + \frac{1}{3}, \quad a_y = -3\lambda - \frac{2}{3}, \quad a_z = 5\lambda - \frac{2}{3}
\]

Now, squaring and summing:

\[
(\lambda + \frac{1}{3})^2 + (-3\lambda - \frac{2}{3})^2 + (5\lambda - \frac{2}{3})^2 = 1
\]

Expanding:

\[
(\lambda^2 + \frac{2}{3}\lambda + \frac{1}{9}) + (9\lambda^2 + 4\lambda + \frac{4}{9}) + (25\lambda^2 - \frac{20}{3}\lambda + \frac{4}{9}) = 1
\]

\[
\lambda^2 + 9\lambda^2 + 25\lambda^2 + \frac{2}{3}\lambda + 4\lambda - \frac{20}{3}\lambda + \frac{1}{9} + \frac{4}{9} + \frac{4}{9} = 1
\]

\[
35\lambda^2 - \frac{14}{3}\lambda + \frac{9}{9} = 1
\]

\[
35\lambda^2 - \frac{14}{3}\lambda + 1 = 1
\]

\[
35\lambda^2 - \frac{14}{3}\lambda = 0
\]

\[
35\lambda (\lambda - \frac{2}{15}) = 0
\]

### Step 5: Solving for λ
\[
\lambda = 0 \quad \text{or} \quad \lambda = \frac{2}{15}
\]

### Step 6: Finding a
For **λ = 2/15**:

\[
a_x = \frac{2}{15} + \frac{1}{3} = \frac{2}{15} + \frac{5}{15} = \frac{7}{15}
\]

\[
a_y = -3 \times \frac{2}{15} - \frac{2}{3} = -\frac{6}{15} - \frac{10}{15} = -\frac{16}{15}
\]

\[
a_z = 5 \times \frac{2}{15} - \frac{2}{3} = \frac{10}{15} - \frac{10}{15} = 0
\]

Thus, **a** is:

\[
a = \frac{7}{15} i - \frac{16}{15} j
\]

This satisfies the unit vector condition.

### Final Answer:
\[
a = \frac{7}{15} i - \frac{16}{15} j
\]