Let vec{a} and vec{b} be two non-collinear ve | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the equation $2\vec{u} - \vec{v} = \vec{w}$.Substituting the expressions for $\vec{u}$,$\vec{v}$,and $\vec{w}$:$2(x\vec{a} + 2y\vec{b}) - (-

Vedclass · Accepted Answer

$x = 10/7, y = 4/7$

Vedclass · Answer

(B) Given the equation $2\vec{u} - \vec{v} = \vec{w}$.Substituting the expressions for $\vec{u}$,$\vec{v}$,and $\vec{w}$:$2(x\vec{a} + 2y\vec{b}) - (-2y\vec{a} + 3x\vec{b}) = 4\vec{a} - 2\vec{b}$$(2x\vec{a} + 4y\vec{b}) + (2y\vec{a} - 3x\vec{b}) = 4\vec{a} - 2\vec{b}$$(2x + 2y)\vec{a} + (4y - 3x)\vec{b} = 4\vec{a} - 2\vec{b}$Since $\vec{a}$ and $\vec{b}$ are non-collinear,we can equate the coefficients:$2x + 2y = 4 \Rightarrow x + y = 2$ (Equation $1$)$-3x + 4y = -2$ (Equation $2$)From Equation $1$,$x = 2 - y$. Substituting into Equation $2$:$-3(2 - y) + 4y = -2$$-6 + 3y + 4y = -2$$7y = 4 \Rightarrow y = 4/7$Substituting $y = 4/7$ into $x = 2 - y$:$x = 2 - 4/7 = 10/7$Thus,$x = 10/7$ and $y = 4/7$.

Let $\vec{a}$ and $\vec{b}$ be two non-collinear vectors. For what values of $x$ and $y$ is the equation $2\vec{u} - \vec{v} = \vec{w}$ true,where $\vec{u} = x\vec{a} + 2y\vec{b}$,$\vec{v} = -2y\vec{a} + 3x\vec{b}$,and $\vec{w} = 4\vec{a} - 2\vec{b}$?

Similar Questions

If the position vectors of the points $A$ and $B$ are $\vec{a} = \hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} - 3\hat{k}$,then what will be the position vector of the midpoint of $AB$?

$(a \cdot i)i + (a \cdot j)j + (a \cdot k)k = $

If $a, b, c$ are non-collinear vectors such that for some scalars $x, y, z,$ $xa + yb + zc = 0,$ then

Let $\vec{u}, \vec{v}, \vec{w}$ be vectors such that $\vec{u} + \vec{v} + \vec{w} = \vec{0}$. If $|\vec{u}| = 3$,$|\vec{v}| = 4$,and $|\vec{w}| = 5$,then $\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}$ is:

In a right-angled triangle,if the position vector of the vertex having the right angle is $\vec{A} = -3\hat{i} + 5\hat{j} + 2\hat{k}$ and the position vector of the midpoint of its hypotenuse is $\vec{M} = 6\hat{i} + 2\hat{j} + 5\hat{k}$,then the position vector of its centroid is

Vedclass Test Series

Exam Paper Generator

Online Exam Module