(B) Given the equation of a line passing through the points with position vectors $a$ and $b$ is $r = ta + (1-t)b$,where $t$ is a scalar parameter.Rea

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a point on the line joining the points whose position vectors are $a$ and $b$

(B) Given the equation of a line passing through the points with position vectors $a$ and $b$ is $r = ta + (1-t)b$,where $t$ is a scalar parameter.Rearranging the equation,we get $r - b = t(a - b)$.This implies that the vector $(r - b)$ is collinear with the vector $(a - b)$.Thus,the locus of the point represented by $r$ is the straight line passing through the points represented by vectors $a$ and $b$.When $0 \leq t \leq 1$,the point $r$ lies on the line segment joining the points with position vectors $a$ and $b$.

Class 12 Mathematics Vector Algebra Basic , Modulus and Algebra of vectors

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If $a$ and $b$ represent two non-collinear vectors,the equation $r = ta + (1-t)b$ represents

TS EAMCET 2020 All TS EAMCET

A
a point on the third side of a triangle for which $a$ and $b$ are two sides,only when $0 \leq t \leq 1$
B
a point on the line joining the points whose position vectors are $a$ and $b$
C
a vector in the plane of $a$ and $b$ only when $t > 1$
D
a vector in the plane parallel to the plane of $a$ and $b$,only when $-1 \leq t \leq 1$

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Basic , Modulus and Algebra of vectors

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Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c} = 5$ and $|\vec{a} + \vec{b} - \vec{c}|^2 + |\vec{b} + \vec{c} - \vec{a}|^2 + |\vec{c} + \vec{a} - \vec{b}|^2 = 50$. Then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = $

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If $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k},$ find the unit vector in the direction of
$(i)$ $6 \vec{b}$
(ii) $2 \vec{a}-\vec{b}$

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Vector $\vec{x}$ is a vector in the direction of $(2, -2, 1)$ and has a magnitude of $6$ units. Vector $\vec{y}$ is a vector in the direction of $(1, 1, -1)$ and has a magnitude of $\sqrt{3}$ units. Then,$|\vec{x} + 2\vec{y}| = $ . . . . . . .

EasyGUJCET 2019

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The vectors $\overrightarrow{AB} = 3 \hat{i} + 4 \hat{k}$ and $\overrightarrow{AC} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle $ABC$. The length of the median through $A$ is

EasyMHT CET 2021

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If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{3}$ and $(\vec{a}+\vec{b}+\vec{c})^2+(\vec{b}+\vec{c}-\vec{a})^2+(\vec{c}+\vec{a}-\vec{b})^2=36$,then $|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2=$

EasyAP EAMCET 2024

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If $a$ and $b$ represent two non-collinear vectors,the equation $r = ta + (1-t)b$ represents

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If $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k},$ find the unit vector in the direction of$(i)$ $6 \vec{b}$(ii) $2 \vec{a}-\vec{b}$

Vector $\vec{x}$ is a vector in the direction of $(2, -2, 1)$ and has a magnitude of $6$ units. Vector $\vec{y}$ is a vector in the direction of $(1, 1, -1)$ and has a magnitude of $\sqrt{3}$ units. Then,$|\vec{x} + 2\vec{y}| = $ . . . . . . .

The vectors $\overrightarrow{AB} = 3 \hat{i} + 4 \hat{k}$ and $\overrightarrow{AC} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle $ABC$. The length of the median through $A$ is

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{3}$ and $(\vec{a}+\vec{b}+\vec{c})^2+(\vec{b}+\vec{c}-\vec{a})^2+(\vec{c}+\vec{a}-\vec{b})^2=36$,then $|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2=$

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If $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k},$ find the unit vector in the direction of
$(i)$ $6 \vec{b}$
(ii) $2 \vec{a}-\vec{b}$