Show that the four points A, B, C and D with position vectors $4^i + 5^j +^k, -^j -^k, 3^i + 9^j + 4^k$ and $4 (-^i +^j +^k)$ respectively are coplannar.

The scalar product of the vector $\to a =^i +^j +^k$ with a unit vector along the sum of vector $\to b = 2^i + 4^j + 5^k$ and $\to c = λ^i + 2^j + 3^k$ is equal to one. Find the value of $λ$ and hence find the unit vector along $\to b + \to c$ .

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A vector A is along the positive z−axis and its vector product with another vector B is zero, then vector B could be:

The correct option is C −7^k→A=A^k→A⋅→B=0⇒ for suitable multiply which makes product =0, necessary =−7^k→B=−7^k

A vector $A$ is along the positive $z -$ axis and its vector product with another vector $B$ is zero, then vector $B$ could be:

The correct option is C $- 7^k$
$\to A = A^k$
$\to A \cdot \to B = 0$
$\Rightarrow$ for suitable multiply which makes product $= 0$ , necessary $= - 7^k$
$\to B = - 7^k$