Question

value of n so that vector 2i+3j-2k, 5i+nj+k and -i+2j+3k may lie in same plane

Answer 1

Given: Three vectors 2i+3j-2k, 5i+nj+k and -i+2j+3k

To find: Value of n?

Solution:

• Now we have given that the three vectors lie in same plane, which means vectors are coplanar.

• So for coplanarity, box of a ,b , c should be 0, that means:

            [ a   b   c] = 0

            a . (b x c) = 0

• Let a = 2i+3j-2k, b = 5i+nj+k and c = -i+2j+3k

           [2i+3j-2k    5i+nj+k    -i+2j+3k]  =  2i+3j-2k .  ( 5i+nj+k   x   -i+2j+3k ) = 0

• Now for cross product, we have:

            5i+nj+k   x   -i+2j+3k = $\left[\begin{array}{ccc}i&j&k\\5&n&1\\-1&2&3\end{array}\right]$

                                              = i(3n-2) - j(15 + 1) + k(10 + n)

            5i+nj+k   x   -i+2j+3k = (3n-2)i - 16j + (10+n)k

• Now:

            2i+3j-2k .  ( 5i+nj+k   x   -i+2j+3k )

            2i+3j-2k .  (3n-2)i - 16j + (10+n)k = 0

            6n - 4 - 48 - 20 - 2n = 0

            4n = 72

            n = 18

Answer:

           So the value of n is 18.

Answer 2

Answer:

n=18

Explanation:

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