If vec{a} = t vec{b} where t

Question

(D) Given $\vec{a} = t \vec{b}$ where $t < 0$ is a scalar. Since $t$ is negative,the vectors $\vec{a}$ and $\vec{b}$ must point in opposite directions

Vedclass · Accepted Answer

$\vec{a}, \vec{b}$ are unlike vectors and either $|\vec{a}| \geq |\vec{b}|$ or $|\vec{a}| < |\vec{b}|$

Vedclass · Answer

(D) Given $\vec{a} = t \vec{b}$ where $t Since $t$ is negative,the vectors $\vec{a}$ and $\vec{b}$ must point in opposite directions,which means they are unlike vectors. Taking the modulus on both sides: $|\vec{a}| = |t \vec{b}| = |t| |\vec{b}|$. Since $t If $|t| > 1$,then $|\vec{a}| > |\vec{b}|$. If $|t| = 1$,then $|\vec{a}| = |\vec{b}|$. If $0 Therefore,$\vec{a}$ and $\vec{b}$ are unlike vectors,and the relationship between their magnitudes depends on the value of $|t|$,which can be either $|\vec{a}| \geq |\vec{b}|$ or $|\vec{a}| < |\vec{b}|$.

If $\vec{a} = t \vec{b}$ where $t < 0$ is a scalar,then

Similar Questions

If $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$,then under what condition are the three points $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ collinear?

If $ABCD$ is a parallelogram,$\overrightarrow{AB} = 2i + 4j - 5k$ and $\overrightarrow{AD} = i + 2j + 3k,$ then the unit vector in the direction of $\overrightarrow{BD}$ is

If the vector $a = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and $b$ are collinear and $|b| = 21$, then $b$ is equal to:

Let $\bar{a}$ and $\bar{b}$ be the position vectors of points $A$ and $B$ respectively. $C$ and $D$ are points on the line $AB$ such that $\overline{AC} = 3 \overline{AB}$ and $\overline{BD} = 2 \overline{BA}$. Find the vector $\overline{CD}$.

In $\triangle PQR$,$(4 \hat{i}+3 \hat{j}+6 \hat{k})$,$(2 \hat{i}+2 \hat{j}+3 \hat{k})$ and $(3 \hat{i}+\hat{j}+3 \hat{k})$ are the position vectors of the vertices $P, Q$ and $R$ respectively. Then the position vector of the point of intersection of the angle bisector of $P$ with $QR$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module