Solveeit Logo
Login

Question

Question: If \(a = i + j + k,b = 4i + 3j + 4k\) and \(c = i + \alpha j + \beta k\) are linearly dependent vect...

If a=i+j+k,b=4i+3j+4ka = i + j + k,b = 4i + 3j + 4k and c=i+αj+βkc = i + \alpha j + \beta k are linearly dependent vectors and ∣c∣=3\left| c \right| = \sqrt 3 , then
A) α=1,β=−1\alpha = 1,\beta = - 1
B) α=1,β=±1\alpha = 1,\beta = \pm 1
C) α=−1,β=±1\alpha = - 1,\beta = \pm 1
D) α=±1,β=1\alpha = \pm 1,\beta = 1

Explanation

Solution

In the question it is given that the all the vectors are linearly dependent its mean that the vector triple product of all three vectors is zero that is [a⃗ b⃗ c⃗]=0\left[ {{{\vec a \, \vec b \, \vec c}}} \right] = 0 hence from here we will find the value of β\beta and by using the condition ∣c∣=3\left| c \right| = \sqrt 3 we will find the value of α\alpha .

Complete step by step solution:
In the question it is given that the a=i+j+k,b=4i+3j+4ka = i + j + k,b = 4i + 3j + 4k and c=i+αj+βkc = i + \alpha j + \beta k are linearly dependent vectors ,
It mean that the triple product of the vector is equal to zero
[a⃗ b⃗ c⃗]=0\left[ {{{\vec a \, \vec b \, \vec c}}} \right] = 0
hence from the question
a=i+j+ka = i + j + k
b=4i+3j+4kb = 4i + 3j + 4k
and
c=i+αj+βkc = i + \alpha j + \beta k
So from triple product of this is
[a⃗ b⃗ c⃗]=\left[ {{{\vec a \, \vec b \, \vec c}}} \right] = \left| {\begin{array}{*{20}{l}} 1&1&1 \\\ 4&3&4 \\\ 1&\alpha &\beta \end{array}} \right| =0 = 0
So it is given that the it is equal to zero , from there is at least one row or columns which is equal to zero
By applying operation C2→C2−C1{C_2} \to {C_2} - {C_1} and C3→C3−C1{C_3} \to {C_3} - {C_1}
we get

1&0&0 \\\ 4&{ - 1}&0 \\\ 1&{\alpha - 1}&{\beta - 1} \end{array}} \right|$$$ = 0$ So in the columns $3{\text{rd}}$ all are zero except $\beta - 1$ hence for the determinant equal to zero , $\beta - 1 = 0$ hence $\beta = 1$ It is also given that $\left| c \right| = \sqrt 3 $ mean that the magnitude of vector c is equal to $\sqrt 3 $ $c = i + \alpha j + \beta k$ $\left| c \right| = \sqrt {{1^2} + {\alpha ^2} + {\beta ^2}} = \sqrt 3 $ Squaring both side and put the value $\beta = 1$ $1 + {\alpha ^2} + 1 = 3$ on solving ${\alpha ^2} = 1$ hence $\alpha = \pm 1$ **$\alpha = \pm 1$ $\beta = 1$ hence option D is the correct answer.** **Note:** In vector space if a set of vectors is said to be linearly dependent if at least one of the vectors in the set can be defined as a linear combination of the others and if no vector in the set can be written in this way, then the vectors are said to be linearly independent . Two or more vectors having the same initial point are called coinitial vectors.