MathematicsQ1–2 of 2 questions
Page 1 of 1 · English
1
MathematicsDifficultMCQIIT JEE · 1961
If the line segment joining the points $A(a, b)$ and $B(c, d)$ subtends an angle $\theta$ at the origin,then $\cos \theta$ is equal to
A
$\frac{ab + cd}{\sqrt{(a^2 + b^2)(c^2 + d^2)}}$
B
$\frac{ac + bd}{\sqrt{(a^2 + b^2)(c^2 + d^2)}}$
C
$\frac{ac - bd}{\sqrt{(a^2 + b^2)(c^2 + d^2)}}$
D
None of these
Solution
(B) Let $O$ be the origin $(0, 0)$. The points are $A(a, b)$ and $B(c, d)$.
Using the distance formula,we have:
$(OA)^2 = a^2 + b^2$
$(OB)^2 = c^2 + d^2$
$(AB)^2 = (a - c)^2 + (b - d)^2$
In $\triangle AOB$,by the Law of Cosines:
$\cos \theta = \frac{(OA)^2 + (OB)^2 - (AB)^2}{2(OA)(OB)}$
Substituting the values:
$\cos \theta = \frac{(a^2 + b^2) + (c^2 + d^2) - [(a - c)^2 + (b - d)^2]}{2\sqrt{a^2 + b^2}\sqrt{c^2 + d^2}}$
$\cos \theta = \frac{a^2 + b^2 + c^2 + d^2 - (a^2 - 2ac + c^2 + b^2 - 2bd + d^2)}{2\sqrt{a^2 + b^2}\sqrt{c^2 + d^2}}$
$\cos \theta = \frac{2ac + 2bd}{2\sqrt{a^2 + b^2}\sqrt{c^2 + d^2}}$
$\cos \theta = \frac{ac + bd}{\sqrt{(a^2 + b^2)(c^2 + d^2)}}$
Thus,the correct option is $B$.
Using the distance formula,we have:
$(OA)^2 = a^2 + b^2$
$(OB)^2 = c^2 + d^2$
$(AB)^2 = (a - c)^2 + (b - d)^2$
In $\triangle AOB$,by the Law of Cosines:
$\cos \theta = \frac{(OA)^2 + (OB)^2 - (AB)^2}{2(OA)(OB)}$
Substituting the values:
$\cos \theta = \frac{(a^2 + b^2) + (c^2 + d^2) - [(a - c)^2 + (b - d)^2]}{2\sqrt{a^2 + b^2}\sqrt{c^2 + d^2}}$
$\cos \theta = \frac{a^2 + b^2 + c^2 + d^2 - (a^2 - 2ac + c^2 + b^2 - 2bd + d^2)}{2\sqrt{a^2 + b^2}\sqrt{c^2 + d^2}}$
$\cos \theta = \frac{2ac + 2bd}{2\sqrt{a^2 + b^2}\sqrt{c^2 + d^2}}$
$\cos \theta = \frac{ac + bd}{\sqrt{(a^2 + b^2)(c^2 + d^2)}}$
Thus,the correct option is $B$.
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2
MathematicsDifficultMCQIIT JEE · 1961
If $O$ is the origin and the coordinates of any two points $Q_1$ and $Q_2$ are $(x_1, y_1)$ and $(x_2, y_2)$ respectively,then $OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2 = $
A
$x_1x_2 - y_1y_2$
B
$x_1y_1 - x_2y_2$
C
$x_1x_2 + y_1y_2$
D
$x_1y_1 + x_2y_2$
Solution
(C) From triangle $OQ_1Q_2$,by applying the cosine rule:
$Q_1Q_2^2 = OQ_1^2 + OQ_2^2 - 2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
Using the distance formula,$Q_1Q_2^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2$,$OQ_1^2 = x_1^2 + y_1^2$,and $OQ_2^2 = x_2^2 + y_2^2$.
Substituting these into the cosine rule:
$(x_1 - x_2)^2 + (y_1 - y_2)^2 = (x_1^2 + y_1^2) + (x_2^2 + y_2^2) - 2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
$x_1^2 - 2x_1x_2 + x_2^2 + y_1^2 - 2y_1y_2 + y_2^2 = x_1^2 + y_1^2 + x_2^2 + y_2^2 - 2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
$-2x_1x_2 - 2y_1y_2 = -2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
Dividing by $-2$,we get:
$OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2 = x_1x_2 + y_1y_2$
$Q_1Q_2^2 = OQ_1^2 + OQ_2^2 - 2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
Using the distance formula,$Q_1Q_2^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2$,$OQ_1^2 = x_1^2 + y_1^2$,and $OQ_2^2 = x_2^2 + y_2^2$.
Substituting these into the cosine rule:
$(x_1 - x_2)^2 + (y_1 - y_2)^2 = (x_1^2 + y_1^2) + (x_2^2 + y_2^2) - 2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
$x_1^2 - 2x_1x_2 + x_2^2 + y_1^2 - 2y_1y_2 + y_2^2 = x_1^2 + y_1^2 + x_2^2 + y_2^2 - 2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
$-2x_1x_2 - 2y_1y_2 = -2OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2$
Dividing by $-2$,we get:
$OQ_1 \cdot OQ_2 \cos \angle Q_1OQ_2 = x_1x_2 + y_1y_2$
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