Answer:
(D) 2
Step-by-step explanation:
If two or more vectors are linearly dependent, then one of them can be expressed as a linear combination of the rest. In other words, vectors are dependent linearly on one another if the determinant of the matrix that they form is zero.
Given vectors:
i+j+2k
i+pj+5k
5i+3j+4k
Now, since they are linearly dependent, then the determinant of their matrix should be zero. i.e
Let their matrix be A and is given by;
A = ![\left[\begin{array}{ccc}1&1&2\\1&p&5\\5&3&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%262%5C%5C1%26p%265%5C%5C5%263%264%5Cend%7Barray%7D%5Cright%5D)
Where;
the first column holds the i components of the vectors
the second column holds the j components of the vectors and
the third column holds the k components of the vectors
=> |A| = det(A) = 0
![det(A) = det \left[\begin{array}{ccc}1&1&2\\1&p&5\\5&3&4\end{array}\right] = 0](https://tex.z-dn.net/?f=det%28A%29%20%3D%20det%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%262%5C%5C1%26p%265%5C%5C5%263%264%5Cend%7Barray%7D%5Cright%5D%20%3D%200)
<em>Now, let's calculate the determinant.</em>
<em></em>
![det \left[\begin{array}{ccc}1&1&2\\1&p&5\\5&3&4\end{array}\right] = 1(4p - 15) - 1(4 - 25) + 2(3 - 5p) = 0](https://tex.z-dn.net/?f=det%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%262%5C%5C1%26p%265%5C%5C5%263%264%5Cend%7Barray%7D%5Cright%5D%20%3D%201%284p%20-%2015%29%20-%201%284%20-%2025%29%20%2B%202%283%20-%205p%29%20%3D%200)
=> 4p -15 - 4 + 25 + 6 - 10p = 0
=> 4p - 10p + 12 = 0
=> 6p = 12
=> p = 2
Therefore, the value of p for which the vectors are linearly dependent is 2