Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Three-dimensional Geometry

Question:

If $l_1, m_1, n_1; l_2, m_2, n_2$ and $l_3, m_3, n_3$ are direction cosines of three mutually perpendicular lines then, the value of $\begin{vmatrix}l_1& m_1& n_1\\l_2& m_2& n_2\\l_3& m_3& n_3\end{vmatrix}$, is 

Options:

$l_3\, m_3\, n_3$

$±1$

$l_1\, m_1\, n_1$

$l_2\, m_2\, n_2$

Correct Answer:

$±1$

Explanation:

Since $l_1, m_1, n_1; l_2, m_2, n_2$ and $l_3, m_3, n_3$ are direction cosines of three mutually perpendicular lines

$∴l_1^2+m_1^2+n_1^2=1,l_2^2+m_2^2+n_2^2=1, l_3^2+m_3^2+n_3^2=1, l_1l_2+m_1m_2+n_1n_2=0,l_1l_3+m_1m_3+n_1n_3=0,l_2l_3+m_2m_3+n_2n_3=0$

Let $Δ=\begin{vmatrix}l_1& m_1& n_1\\l_2& m_2& n_2\\l_3& m_3& n_3\end{vmatrix}$. Then,

$Δ^2=\begin{vmatrix}l_1& m_1& n_1\\l_2& m_2& n_2\\l_3& m_3& n_3\end{vmatrix}\begin{vmatrix}l_1& m_1& n_1\\l_2& m_2& n_2\\l_3& m_3& n_3\end{vmatrix}$

$=\begin{vmatrix}l_1^2+m_1^2+n_1^2&l_1l_2+m_1m_2+n_1n_2&l_1l_3+m_1m_3+n_1n_3\\l_1l_2+m_1m_2+n_1n_2&l_2^2+m_2^2+n_2^2&l_2l_3+m_2m_3+n_2n_3\\l_1l_3+m_1m_3+n_1n_3&l_2l_3+m_2m_3+n_2n_3&l_3^2+m_3^2+n_3^2\end{vmatrix}$

$=\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}=1$

$∴Δ=±1$