JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 11)
Let L1 and L2 be the following straight lines.
$${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$$ and $${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$$.
Suppose the straight line
$$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$$
lies in the plane containing L1 and L2 and passes through the point of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE?
$${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$$ and $${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$$.
Suppose the straight line
$$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$$
lies in the plane containing L1 and L2 and passes through the point of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE?
$$\alpha $$ $$-$$ $$\gamma $$ = 3
l + m = 2
$$\alpha $$ $$-$$ $$\gamma $$ = 1
l + m = 0
Spiegazione
Equation of given straight lines
$${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$$
and $${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$$
having vector form respectively, are
and $$\eqalign{ & r = (\widehat i + \widehat k) + \lambda (\widehat i - \widehat j + 3\widehat k) \cr & r = (\widehat i + \widehat k) + v( - 3\widehat i - \widehat j + \widehat k) \cr} $$
$$ \because $$ $$(\widehat i - \widehat j + 3\widehat k)\,.\,( - 3\widehat i - \widehat j + \widehat k)$$
= $$-$$3 + 1 + 3 = 1 is positive,
$$ \therefore $$ Angle between supporting line vectors of lines L1 and L2 is acute, and point of intersection of given lines L1 and L2 is (1, 0, 1)
Now, vector along the acute angle bisector of vectors $$(\widehat i - \widehat j + 3\widehat k)$$ and $$( - 3\widehat i - \widehat j + \widehat k)$$ is $$(\widehat i - \widehat j + 2\widehat k)$$ or $$(\widehat i + \widehat j - 2\widehat k)$$
It is given that line
$$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$$ is the bisector of the acute angle between the lines L1 and L2, so
l = 1 and m = 1
and $${{I - \alpha } \over 1} = {{0 - 1} \over 1} = {{1 - \gamma } \over { - 2}}$$
$$ \Rightarrow \alpha = 2,\,\gamma = - 1$$
$$ \therefore $$ $$\alpha - \gamma = 3,\,l + m = 2$$
$${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$$
and $${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$$
having vector form respectively, are
and $$\eqalign{ & r = (\widehat i + \widehat k) + \lambda (\widehat i - \widehat j + 3\widehat k) \cr & r = (\widehat i + \widehat k) + v( - 3\widehat i - \widehat j + \widehat k) \cr} $$
$$ \because $$ $$(\widehat i - \widehat j + 3\widehat k)\,.\,( - 3\widehat i - \widehat j + \widehat k)$$
= $$-$$3 + 1 + 3 = 1 is positive,
$$ \therefore $$ Angle between supporting line vectors of lines L1 and L2 is acute, and point of intersection of given lines L1 and L2 is (1, 0, 1)
Now, vector along the acute angle bisector of vectors $$(\widehat i - \widehat j + 3\widehat k)$$ and $$( - 3\widehat i - \widehat j + \widehat k)$$ is $$(\widehat i - \widehat j + 2\widehat k)$$ or $$(\widehat i + \widehat j - 2\widehat k)$$
It is given that line
$$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$$ is the bisector of the acute angle between the lines L1 and L2, so
l = 1 and m = 1
and $${{I - \alpha } \over 1} = {{0 - 1} \over 1} = {{1 - \gamma } \over { - 2}}$$
$$ \Rightarrow \alpha = 2,\,\gamma = - 1$$
$$ \therefore $$ $$\alpha - \gamma = 3,\,l + m = 2$$