Question

Statement-1 : If a line makes acute angles $\alpha ,\beta ,\gamma ,\delta$ with diagonals of a cube, then ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta =\frac{4}{3}$
Statement 2 : If a line makes equal angle (acute) with the axes, then its direction cosine are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{3}}$

• A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
• B. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
• C. Statement-1 is True, Statement-2 is False
• D. Statement-1 is False, Statement-2 is True

Answer 1

Answer: (C)

Statement-1 is True, Statement-2 is False

Let l,m,n be the direction ratio of a line making angles ,,, with four diagonals of a cube then as four diagonals have direction ratio,$(a,a,a)$;$(a,a,-a)$;$(a,-a,a,)$ ;$(-a,a,a)$(where $a$ is the side of the cube) Now, $\cos \alpha =\left|\frac{al+am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}}\right|=\left|\frac{l+m+n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}\right|$ $\cos \beta =\left|\frac{al+am-an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}}\right|=\left|\frac{l+m-n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}\right|$ $\cos \gamma =\left|\frac{al-am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}}\right|=\left|\frac{l-m+n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}\right|$ $\cos \delta =\left|\frac{al+am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}}\right|=\left|\frac{-l+m+n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}\right|$ and ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta$ $=\frac{1}{3}\times \frac{(4l^2+4m^2+4n^2)}{l^2+m^2+n^2}=\frac{4}{3}$ So statement 1 is true and statement-2 is not true . Hence, option 'C' is correct.