JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 7)
If a point R(4, y, z) lies on the line segment joining
the points P(2, –3, 4) and Q(8, 0, 10), then the
distance of R from the origin is :
$$2 \sqrt {14}$$
$$ \sqrt {53}$$
$$2 \sqrt {21}$$
6
Explanation
Equation of PQ is
$${{x - 2} \over {8 - 2}} = {{y + 3} \over {0 - \left( { - 3} \right)}} = {{z - 4} \over {10 - 4}}$$
$$ \Rightarrow $$ $${{x - 2} \over 6} = {{y + 3} \over 3} = {{z - 4} \over 6}$$
Point R (4, y, z) lies on this
$$ \therefore $$ $${{4 - 2} \over 6} = {{y + 3} \over 3} = {{z - 4} \over 6}$$
$$ \Rightarrow $$ $${1 \over 3} = {{y + 3} \over 3} = {{z - 4} \over 6}$$
y = -2 and y = 6
$$ \therefore $$ R = (4, -2, 6)
Distance of R(4, -2, 6) from the origin O(0, 0, 0) is
RO = $$\sqrt {{4^2} + {{\left( { - 2} \right)}^2} + {6^2}} $$
= $$\sqrt {56} = 2\sqrt {14} $$
$${{x - 2} \over {8 - 2}} = {{y + 3} \over {0 - \left( { - 3} \right)}} = {{z - 4} \over {10 - 4}}$$
$$ \Rightarrow $$ $${{x - 2} \over 6} = {{y + 3} \over 3} = {{z - 4} \over 6}$$
Point R (4, y, z) lies on this
$$ \therefore $$ $${{4 - 2} \over 6} = {{y + 3} \over 3} = {{z - 4} \over 6}$$
$$ \Rightarrow $$ $${1 \over 3} = {{y + 3} \over 3} = {{z - 4} \over 6}$$
y = -2 and y = 6
$$ \therefore $$ R = (4, -2, 6)
Distance of R(4, -2, 6) from the origin O(0, 0, 0) is
RO = $$\sqrt {{4^2} + {{\left( { - 2} \right)}^2} + {6^2}} $$
= $$\sqrt {56} = 2\sqrt {14} $$
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