Question

If $a = i + j + k,b = 4i + 3j + 4k$ and $c = i + \alpha j + \beta k$ are linearly dependent vectors and $\left| c \right| = \sqrt 3 $ , then
A) $\alpha = 1,\beta = - 1$
B) $\alpha = 1,\beta = \pm 1$
C) $\alpha = - 1,\beta = \pm 1$
D) $\alpha = \pm 1,\beta = 1$

Answer 1

Hint:
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 $\left[ {{{\vec a \, \vec b \, \vec c}}} \right] = 0$ hence from here we will find the value of $\beta $ and by using the condition $\left| c \right| = \sqrt 3 $ we will find the value of $\alpha $.

Complete step by step solution:
In the question it is given that the $a = i + j + k,b = 4i + 3j + 4k$ and $c = i + \alpha j + \beta k$ are linearly dependent vectors ,
It mean that the triple product of the vector is equal to zero
$\left[ {{{\vec a \, \vec b \, \vec c}}} \right] = 0$
hence from the question
 $a = i + j + k$
$b = 4i + 3j + 4k$
 and
$c = i + \alpha j + \beta k$
So from triple product of this is
$\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$
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 ${C_2} \to {C_2} - {C_1}$ and ${C_3} \to {C_3} - {C_1}$
we get
\[\left| {\begin{array}{*{20}{l}}
  1&0&0 \\
  4&{ - 1}&0 \\
  1&{\alpha - 1}&{\beta - 1}
\end{array}} \right|\]$ = 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.