JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 15)
Let A (3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G
divides BM in the ratio, 2 : 1, then cos ($$\angle $$GOA) (O being the origin) is equal to :
$${1 \over {\sqrt {15} }}$$
$${1 \over {6\sqrt {10} }}$$
$${1 \over {\sqrt {30} }}$$
$${1 \over {2\sqrt {15} }}$$
Explanation
G is the centroid of $$\Delta $$ABC
_10th_April_Morning_Slot_en_15_2.png)
$$OG = \sqrt {4 + 16 + 4} $$, OA = $$\sqrt {9 + 1} $$
$$AG = \sqrt {1 + 16 + 9} $$
$$\cos \theta = {{24 + 10 - 26} \over {2\sqrt {24} \sqrt {10} }}$$
$$ \Rightarrow {8 \over {2\sqrt {8 \times 3 \times 2 \times 5} }}$$
$$ \Rightarrow {4 \over {4\sqrt {15} }} = {1 \over {\sqrt {15} }}$$
$$OG = \sqrt {4 + 16 + 4} $$, OA = $$\sqrt {9 + 1} $$
$$AG = \sqrt {1 + 16 + 9} $$
$$\cos \theta = {{24 + 10 - 26} \over {2\sqrt {24} \sqrt {10} }}$$
$$ \Rightarrow {8 \over {2\sqrt {8 \times 3 \times 2 \times 5} }}$$
$$ \Rightarrow {4 \over {4\sqrt {15} }} = {1 \over {\sqrt {15} }}$$
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