Consider the vectors $u = a \hat{i} + b \hat{j} + c \hat{k}$,$v = a^2 \hat{i} + b^2 \hat{j} + c^2 \hat{k}$ and $w = a^3 \hat{i} + b^3 \hat{j} + c^3 \hat{k}$. These vectors are coplanar if and only if

  • A
    all $a, b$ and $c$ are equal
  • B
    one of $a, b$ and $c$ is zero
  • C
    any two of $a, b$ and $c$ are equal
  • D
    either one of $a, b$ and $c$ is zero,or any two of $a, b$ and $c$ are equal

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Similar Questions

The volume of a parallelepiped,whose coterminous edges are given by $\bar{u}=\hat{i}+\hat{j}+\lambda \hat{k}$,$\bar{v}=\hat{i}+\hat{j}+3 \hat{k}$,and $\bar{w}=2 \hat{i}+\hat{j}+\hat{k}$,is $1$ cubic unit. If $\theta$ is the angle between $\bar{u}$ and $\bar{w}$,then the value of $\cos \theta$ is:

Let $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{B} = \hat{i}$,and $\overrightarrow{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$. If $C_2 = -1$ and $C_3 = 1$,then to make the three vectors coplanar:

If the vectors $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$,$\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}$,and $\vec{c}=3 \hat{i}+p \hat{j}+5 \hat{k}$ are coplanar,then $p=$

The volume of the parallelepiped whose coterminous edges are $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = 3\hat{i} - \hat{j} - \hat{k}$, and $\vec{c} = \hat{j} + 3\hat{k}$ is:

If the four points $A(6,2,4)$,$B(1,3,5)$,$C(1,-2,3)$,and $D(6, k, 2)$ are coplanar,then $k=$

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