If two vectors 2i+3j-k and - 4i-6j-¥k are parallel to each other, then what will be the value of ¥?I I||||Ans:2 ¦¦¦¦¦Plz provide me the soln?

Question

Answer

Answer

Solution:

Two vectors are parallel to each other if their cross product is zero.

So, (2i + 3j - k) x (-4i - 6j - ¥k) = 0

We can expand the cross product using determinants,

| i j k |
| 2 3 -1 |
|-4 -6 -¥ |

= i[(3 x (-¥)) - (-1) x (-6)] - j[(2 x (-¥)) - (-1) x (-4)] + k[(2 x (-6)) - (3 x (-4))]

= i(3¥ + 6) - j(-2¥ + 4) + k(-12 - 12)

= (3¥ + 6)i + (2¥ - 4)j - 24k

For the cross product to be zero, each component must be zero,

3¥ + 6 = 0
2¥ - 4 = 0
-24 = 0 (which is not possible)

Solving the first two equations, we get ¥ = -2 and this is the required answer.

Therefore, the value of ¥ is 2.

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