Given that \(\vec{a}.\vec{b}=0\) and \(\vec{a}\times\vec{b}=0\).What can you conclude about the vectors \(\vec{a}\) and \(\vec{b}\) ?
Show that\((\vec{a}-\vec{b})\times(\vec{a}+\vec{b})=2(\vec{a}\times \vec{b})\)
The cartesian equation of a line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\). Write its vector form.
Find the cartesian equation of the line that passes through the point(-2,4,-5)and parallel to the line given by \(\frac{x+3}{3}\)=\(\frac{y-4}{5}\)=\(\frac{z+8}{6}\)
Find the vector equation of the plane passing through the intersection of the planes
\(\overrightarrow r.(2\hat i+2\hat j-3\hat k)=7\),\(\overrightarrow r.(2\hat i+5\hat j+3\hat k)=9\) and through the point ( 2, 1, 3 ).
Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2\(\hat i\)-\(\hat j\)+4\(\hat k\) and is in the direction \(\hat i\)+2\(\hat j\)-\(\hat k\).
Find the equation of the plane through the intersection of the planes3x-y+2z-4 =0 and x+y+z-2=0 and the point (2, 2, 1).
Find the equation of the line that passes through the point(1,2,3)and is parallel to the vector 3\(\hat i\)+2\(\hat j\)-2\(\hat k\).
Find the intercepts cut off by the plane 2x+y-z = 5
Find the equation of the planes that passes through three points.
(a) (1,1,-1),(6,4,-5),(-4,-2,-3)
(b) (1,1,0),(1,2,1),(-2,2,-1).
Show that the three lines with direction cosines\(\frac{12}{13}\),\(-\frac{3}{13}\),\(-\frac{4}{13}\) ; \(\frac{4}{13}\),\(\frac{12}{13}\),\(\frac{3}{13}\);\(\frac{3}{13}\),\(-\frac{4}{13}\),\(\frac{12}{13}\) are mutually perpendicular.