JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 13)
In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3$$\widehat i$$ + $$\widehat j$$ $$-$$ $$\widehat k$$, $$-$$$$\widehat i$$ + 3$$\widehat j$$ + p$$\widehat k$$ and 5$$\widehat i$$ + q$$\widehat j$$ $$-$$ 4$$\widehat k$$, then the point (p, q) lies
on a line :
parallel to x-axis.
parallel to y-axis.
making an acute angle with the positive direction of x-axis.
making an obtuse angle with the positive direction of x-axis.
Explanation
Given,
$$\overrightarrow A = 3\widehat i + \widehat j - \widehat k$$
$$\overrightarrow B = - \widehat i + 3\widehat j - p\widehat k$$
$$\overrightarrow C = 5\widehat i + 9\widehat j - 4\widehat k$$
$$ \therefore $$ $$\overrightarrow {AB} = - 4\widehat i + 2\widehat j + \left( {p + 1} \right)\widehat k$$
$$\overrightarrow {AC} = 2\widehat i + \left( {q - 1} \right)\widehat j - 3\widehat k$$
_9th_April_Morning_Slot_en_13_1.png)
$$\Delta $$ABC is a right angle triangle.
Here $$\overrightarrow {AB} $$ perpendicular to $$\overrightarrow {AC} $$
$$ \therefore $$ $$\overrightarrow {AB} $$ . $$\overrightarrow {AC} $$ = 0
$$ \Rightarrow $$ $$-$$ 8 + 2(q $$-$$ 1) $$-$$ 3(p + 1) = 0
$$ \Rightarrow $$ 3p $$-$$ 2q + 13 = 0
$$ \therefore $$ (p, q) lies on the line
3x $$-$$ 2y + 13 = 0
And slope of the line = $${3 \over 2}$$
$$ \therefore $$ line makes an angle less than 90o or acute angle with the positive direction of x-axis.
$$\overrightarrow A = 3\widehat i + \widehat j - \widehat k$$
$$\overrightarrow B = - \widehat i + 3\widehat j - p\widehat k$$
$$\overrightarrow C = 5\widehat i + 9\widehat j - 4\widehat k$$
$$ \therefore $$ $$\overrightarrow {AB} = - 4\widehat i + 2\widehat j + \left( {p + 1} \right)\widehat k$$
$$\overrightarrow {AC} = 2\widehat i + \left( {q - 1} \right)\widehat j - 3\widehat k$$
$$\Delta $$ABC is a right angle triangle.
Here $$\overrightarrow {AB} $$ perpendicular to $$\overrightarrow {AC} $$
$$ \therefore $$ $$\overrightarrow {AB} $$ . $$\overrightarrow {AC} $$ = 0
$$ \Rightarrow $$ $$-$$ 8 + 2(q $$-$$ 1) $$-$$ 3(p + 1) = 0
$$ \Rightarrow $$ 3p $$-$$ 2q + 13 = 0
$$ \therefore $$ (p, q) lies on the line
3x $$-$$ 2y + 13 = 0
And slope of the line = $${3 \over 2}$$
$$ \therefore $$ line makes an angle less than 90o or acute angle with the positive direction of x-axis.
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