Show that a vector coplanar with - - - - 2 k- and - - 2 - - k- and perpendicular to - - - - k- is - 1 - 2 - - k- -

A vector of magnitude $\sqrt{2}$ coplanar with the vectors $\to a =^i +^j + 2^k and \to b =^i + 2^j +^k$ and perpendicular to the vector $\to c =^i +^j +^k,$ is

Q. A unit vector in the plane of

^i + 2^j +^k

and

^i +^j + 2^k

and perpendicular to

2^i +^j +^k

, is

Join BYJU'S Learning Program

Show that a vector coplanar with ^i+^j+2^k and ^i+2^j+^k and perpendicular to ^i+^j+^k is ±1√2(^j−^k).

Let →a=ˆi+ˆj+2ˆk→b=ˆi+2ˆj+ˆk→n=ˆi+ˆj+ˆk→c=x→a+y→b→c⋅→n=0=x(→a⋅→n)+y(→b⋅→n)⇒x(4)+y(4)=0y=−x→c=x(→a−→b)=x(−ˆj+ˆk)∣∣→c∣∣=1√x2+x2=1√2x2=12x2=1x2=12x=±1√2→c=±1√2(−ˆj+ˆk)⇒1√2(ˆj−ˆk),1√2(ˆk−ˆj)

Show that a vector coplanar with $^i +^j + 2^k$ and $^i + 2^j +^k$ and perpendicular to $^i +^j +^k$ is $\pm \frac{1}{\sqrt{2}} (^j -^k)$ .