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If the direction cosines of a line are $\left(\frac{a}{\sqrt{83}}, \frac{5}{\sqrt{83}}, \frac{c}{\sqrt{83}}\right)$ and $c-a=4$, then $c a=$
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The correct answer is:
$21$
Direction cosine of $a \hat{i}+b \hat{j}+c \hat{k}$ can be
$\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$
According to question,
$\sqrt{a^2+b^2+c^2}=\sqrt{83}, b=5$
$\Rightarrow \quad a^2+(5)^2+c^2=83$
$\Rightarrow \quad a^2+c^2=83-25$
$\Rightarrow a^2+c^2=58$ ...(i)
and $c-a=4$ [given]
$(c-a)^2+2 a c=a^2+c^2$
$\Rightarrow(4)^2+2 a c=58 \quad$ [from Eq. (i), $\left.c^2+a^2=58\right]$
$\Rightarrow 2 a c=58-16$
$a c=\frac{42}{2}=21$
$\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$
According to question,
$\sqrt{a^2+b^2+c^2}=\sqrt{83}, b=5$
$\Rightarrow \quad a^2+(5)^2+c^2=83$
$\Rightarrow \quad a^2+c^2=83-25$
$\Rightarrow a^2+c^2=58$ ...(i)
and $c-a=4$ [given]
$(c-a)^2+2 a c=a^2+c^2$
$\Rightarrow(4)^2+2 a c=58 \quad$ [from Eq. (i), $\left.c^2+a^2=58\right]$
$\Rightarrow 2 a c=58-16$
$a c=\frac{42}{2}=21$
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