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# The vectors $\overrightarrow{2i}+\overrightarrow{3j}+\overrightarrow{4k}$ and $\overrightarrow{ai}+\overrightarrow{bj}+\overrightarrow{ck}$ are perpendicular when

$\begin{array} {1 1}(1)a=2, b=3 ,c=-4 &(2)a=4 ,b= 4 , c=5\\(3)a=4 ,b = 4 ,c=-5 &(4)a=-2 ,b=3 ,c=4 \end{array}$

Given $\overrightarrow{2i}+\overrightarrow{3j}+\overrightarrow{4k}$ and $\overrightarrow{ai}+\overrightarrow{bj}+\overrightarrow{ck}$ are perpendicular
$\therefore \overrightarrow{2i}+\overrightarrow{3j}+\overrightarrow{4k}.\overrightarrow{ai}+\overrightarrow{bj}+\overrightarrow{ck}=0$
$2a+3+4c=0$ ----(1)
$a=4 ,b = 4 ,c=-5$ is the only values satisfying equation.
Hence 3 is the correct answer.