JEE Main 2003 — Vectors & 3D Geometry

If $\left|\begin{array}{lll}a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3\end{array}\right|=0$ and vectors $\left(1, a, a^2\right),\left(a, b, b^2\right)$ and $\left(a, c, c^2\right)$ are non-coplanar, then the product abc equals

$\vec{a}, \vec{b}, \vec{c}$ are 3 vectors, such that $\vec{a}+\vec{b}+\vec{c}=0,|\vec{a}|=1,|\vec{b}|=2 \mid \vec{a}$ then $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is equal to

The lines $\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-4}{-\mathrm{k}}$ and $\frac{\mathrm{x}-1}{\mathrm{k}}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{1}$ are coplanar if

The resultant of forces $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is $\overrightarrow{\mathrm{R}}$. If $\overrightarrow{\mathrm{Q}}$ is doubled then $\overrightarrow{\mathrm{R}}$ is doubled. If the direction of $\overrightarrow{\mathrm{Q}}$ is reversed, then $\overrightarrow{\mathrm{R}}$ is again doubled. Then $\mathrm{P}^2: \mathrm{Q}^2: \mathrm{R}^2$ is

A particle acted on by constant forces $4 \hat{i}+\hat{j}-3 \hat{k}$ and $3 \hat{i}+\hat{j}-\hat{k}$ to the point $5 \hat{i}+4 \hat{j}-\hat{k}$. The total work done by the forces is

If $\vec{u}, \vec{v}$ and $\vec{w}$ are three non-coplanar vectors, then $(\vec{u}+\vec{v}-\vec{w}) .(\vec{u}-\vec{v}) \times(\vec{v}-\vec{w})$ equals

A couple is of moment $\overrightarrow{\mathrm{G}}$ and the force forming the couple is $\overrightarrow{\mathrm{P}}$. If $\overrightarrow{\mathrm{P}}$ is turned through a right angle the moment of the couple thus formed is $\overrightarrow{\mathrm{H}}$. If instead, the force $\overrightarrow{\mathrm{P}}$ are turned through an angle $\alpha$, then the moment of couple becomes

Let $\overrightarrow{\mathrm{u}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{w}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. If $\hat{\mathrm{n}}$ is a unit vector such that $\vec{u} \cdot \hat{n}=0$ and $\overrightarrow{\mathrm{v}} \cdot \hat{\mathrm{n}}=0$, then $|\overrightarrow{\mathrm{w}} \cdot \hat{\mathrm{n}}|$ is equal to

Consider points A, B, C and D with position vectors $7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}, \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}},-\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and $5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ respectively. Then $\mathrm{ABCD}$ is a

The two lines $x=a y+b, z=c y+d$ and $x=a^{\prime} y+b^{\prime} z=c^{\prime} y+d^{\prime}$ will be perpendicular, if and only if