Q100 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics
Assertion (A) The direction ratios of line $L_1$ are 2, 5, 7 and those of line $L_2$ are $\frac{4}{\sqrt{19}}, \frac{10}{\sqrt{19}}$, $\frac{14}{\sqrt{19}}$. The lines $L_1, L_2$ are parallel.Reason (R) The direction ratios of a line $L_1$ are $a_1, b_1, c_1$ and those of another line $L_2$ are $a_2, b_2, c_2$. The lines $L_1$ and $L_2$ are parallel if $a_1 a_2+b_1 b_2+c_1 c_2=0$
The correct option among the following is
- A(A) is true, $(R)$ is true and $(R)$ is the correct explanation for $(A)$.
- B(A) is true, $(R)$ is true but $(R)$ is not the correct explanation for $(A)$.
- Correct answerC$(A)$ is true but $(R)$ is false.
- D(A) is false but $(R)$ is true.
Solution
Let the direction ratios of a line $L_1$ be $a_1, b_1, c_1$ and those of another line $L_2$ are $a_2, b_2, c_2$ and the angle $\theta$ between them is given by
$\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$ $\ldots(i)$
If both lines are parallel then, their direction ratios must be in equal proportion. i.e.,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Now, from Eq. (i)
$\cos \theta=\frac{(2)\left(\frac{4}{\sqrt{19}}\right)+(5)\left(\frac{10}{\sqrt{19}}\right)+(7)\left(\frac{14}{\sqrt{19}}\right)}{\sqrt{2^2+5^2+7^2} \sqrt{\left(\frac{4}{\sqrt{19}}\right)^2+\left(\frac{10}{\sqrt{19}}\right)^2+\left(\frac{14}{\sqrt{19}}\right)^2}}$
$=\frac{\frac{8}{\sqrt{19}}+\frac{50}{\sqrt{19}}+\frac{98}{\sqrt{19}}}{\sqrt{78} \sqrt{\frac{312}{19}}}=\frac{\frac{106}{\sqrt{19}}}{\frac{78 \times 2}{\sqrt{19}}}=1 \Rightarrow \theta=0$
or $\quad \frac{a_1}{a_2}=\frac{2}{\frac{4}{\sqrt{19}}}=\frac{\sqrt{19}}{2} \Rightarrow \frac{b_1}{b_2}=\frac{5}{\frac{10}{\sqrt{19}}}=\frac{\sqrt{19}}{2}$
$\frac{c_1}{c_2}=\frac{7}{\frac{14}{\sqrt{19}}}=\frac{\sqrt{19}}{2} \quad \because \quad \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Hence, both lines are parallel to each other.
Hence, the correct answer is option: 'C'.