(a) Find the condition that the vectors `vec(a)=k hat(i)+l hat(j)` and `vec(b)=l hat(i)+k hat(j)(k, l ne 0)` are parallel.
(b) Show that the vectors `2hat(i)-3hat(j)+4hat(k)` and `-4hat(i)+6hat(j)-8hat(k)` are collinear.

Answer

### Solution:

#### (a) Finding the Condition for Vectors \( \vec{a} \) and \( \vec{b} \) to be Parallel

Given:
- \( \vec{a} = k \hat{i} + l \hat{j} \)
- \( \vec{b} = l \hat{i} + k \hat{j} \)

Vectors \( \vec{a} \) and \( \vec{b} \) are parallel if there exists a scalar \( \lambda \) such that:
\[ \vec{a} = \lambda \vec{b} \]

This implies:
\[ k \hat{i} + l \hat{j} = \lambda (l \hat{i} + k \hat{j}) \]

This leads to two equations:
1. \( k = \lambda l \) (1)
2. \( l = \lambda k \) (2)

From equation (1), we can express \( \lambda \):
\[ \lambda = \frac{k}{l} \quad (l \neq 0) \]

Substituting \( \lambda \) from equation (1) into equation (2):
\[ l = \frac{k}{l} k \]
\[ l^2 = k^2 \]

Thus, we have:
\[ k^2 - l^2 = 0 \]
Factoring gives:
\[ (k - l)(k + l) = 0 \]

This results in two conditions:
1. \( k = l \)
2. \( k = -l \)

Therefore, the conditions for the vectors \( \vec{a} \) and \( \vec{b} \) to be parallel are:
\[ k = l \quad \text{or} \quad k = -l \]

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#### (b) Showing that the Vectors are Collinear

Given vectors:
- \( \vec{m} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \)
- \( \vec{n} = -4 \hat{i} + 6 \hat{j} - 8 \hat{k} \)

To show that \( \vec{m} \) and \( \vec{n} \) are collinear, we need to check if one vector is a scalar multiple of the other.

We can express \( \vec{n} \) in terms of \( \vec{m} \):
\[ \vec{n} = -2 \vec{m} \]

Calculating:
\[ -2 \vec{m} = -2(2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = -4 \hat{i} + 6 \hat{j} - 8 \hat{k} \]

This shows that:
\[ \vec{n} = -2 \vec{m} \]

Since \( \vec{n} \) is a scalar multiple of \( \vec{m} \), we conclude that the vectors \( \vec{m} \) and \( \vec{n} \) are collinear.

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