Prove that if a plane has the intercepts a, b, c and is at a distance of P units from the origin, then \(\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2} =\frac {1}{p^2}\).
Maximise Z=3x+4ySubject to the constrains:x+y≤4,x≥0,y≥0.
Distance between the two planes: 2x+3y+4z = 4 and 4x+6y+8z = 12 is
\(The\ planes: 2x-y+4z=5 \ and \ 5x-2.5y+10z=6\ are\)
Find the angle between the planes whose vector equations are
\(\overrightarrow r.(2\hat i+2\hat j-3\hat k)=5\) and \(\overrightarrow r.(3\hat i-3\hat j+5\hat k)=3\)
If either \(\vec{a}=0\) or \(\vec{b}=0\),then \(\vec{a}\times\vec{b}=0\). Is the converse true? Justify your answer with an example.
Find the equation of the plane through the line of intersection of the planesx+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z= 0
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\(\hat j\)-\(\hat k\).