Two particles are moving with velocities `v_(1)=hati-thatj+hatk` and `v_(2)=thati+thatj+2hatk m//s` respectively. Time at which they are moving perpendicular to each other is.____________(second)

Answer

To solve the problem of finding the time at which two particles are moving perpendicular to each other, we need to follow these steps:

### Step 1: Understand the condition for perpendicularity
Two vectors are perpendicular when their dot product is zero. Therefore, we need to set the dot product of the two velocity vectors equal to zero.

### Step 2: Write down the given velocities
The velocities of the two particles are given as:
- \( \vec{v_1} = \hat{i} - t\hat{j} + \hat{k} \)
- \( \vec{v_2} = t\hat{i} + t\hat{j} + 2\hat{k} \)

### Step 3: Calculate the dot product
The dot product \( \vec{v_1} \cdot \vec{v_2} \) is calculated as follows:
\[ \vec{v_1} \cdot \vec{v_2} = (1)(t) + (-t)(t) + (1)(2) = t - t^2 + 2 \]

### Step 4: Set the dot product equal to zero
To find the time when the particles are perpendicular, we set the dot product equal to zero:
\[ t - t^2 + 2 = 0 \]
Rearranging gives:
\[ -t^2 + t + 2 = 0 \]
or
\[ t^2 - t - 2 = 0 \]

### Step 5: Solve the quadratic equation
We can solve the quadratic equation \( t^2 - t - 2 = 0 \) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -1, c = -2 \).

### Step 6: Substitute the values into the formula
Substituting the values into the formula gives:
\[ t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2} \]

### Step 7: Calculate the possible values of t
This results in two possible values for \( t \):
1. \( t = \frac{4}{2} = 2 \)
2. \( t = \frac{-2}{2} = -1 \)

### Step 8: Choose the valid solution
Since time cannot be negative, we discard \( t = -1 \). Thus, the time at which the two particles are moving perpendicular to each other is:
\[ t = 2 \text{ seconds} \]

### Final Answer
The time at which the two particles are moving perpendicular to each other is **2 seconds**.

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