AP EAMCET · Maths · Three Dimensional Geometry
Coordinate planes and the planes \(\pi_1, \pi_2, \pi_3\) which are respectively parallel to \(\mathrm{YZ}, \mathrm{ZX}, \mathrm{XY}\) planes at distances \(\mathrm{a}, \mathrm{b}, \mathrm{c}\), form a rectangular parallelpiped, \(\mathrm{d}_1\) is a diagonal of the face on \(\mathrm{XY}\)-plane not passing through origin and \(\mathrm{d}_2\) is diagonal of plane \(\pi_2\) coterminous with \(\mathrm{d}_1\). If none of the coordinates of the vertices of the parallelpiped are negative and angle between \(\mathrm{d}_1\) and \(\mathrm{d}_2\) is \(\theta\), then \(\cos \theta=\)
- A \(\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}\)
- B \(\frac{a}{a^2+b^2+c^2}\)
- C \(\frac{\pi}{2}\)
- D \(\frac{\mathrm{a}^2}{\sqrt{\mathrm{a}^2+\mathrm{b}^2} \sqrt{\mathrm{b}^2+\mathrm{c}^2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}\)
Step-by-step Solution
Detailed explanation
According to question \(\vec{d}_1=a \hat{i}+b \hat{j} \text { and } \vec{d}_2=a \hat{i}+c \hat{j}\) Angle between \(\vec{d}_1\) and \(\vec{d}_2\)…
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