Download App

VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA4 Marks
Question
Express $2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ as the sum of a vector parallel and a vector perpendicular to $2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}.$
✓

Answer

Let $\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\vec{\text{a}}+\vec{\text{b}}\dots(1)$
such that $\vec{\text{a}}$ is a vector parallel to vector $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)$ and $\vec{\text{b}}$ is a vector perpendicular to the vector $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big).$
since, $\vec{\text{a}}$ is parallel to $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)$
$\vec{\text{a}}=\lambda\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)$
$\vec{\text{a}}=2\lambda\hat{\text{i}}+4\lambda\hat{\text{j}}-2\lambda\hat{\text{k}}\dots(2)$
Put value of $\vec{\text{a}}$ in equation (1),
$\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\big(2\lambda\hat{\text{i}}+4\lambda\hat{\text{j}}-2\lambda\hat{\text{k}}\big)+\vec{\text{b}}$
$\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}-2\lambda\hat{\text{j}}-4\lambda\hat{\text{j}}+2\lambda\hat{\text{k}}$
$\vec{\text{b}}=(2-2\lambda)\hat{\text{i}}+(-1-4\lambda)\hat{\text{j}}+(3+2\lambda)\hat{\text{k}}$
$\vec{\text{b}}$ is a vector perpendicuar to the vector $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big),$ then
$\vec{\text{b}}.\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)=0$
$\big[(2-2\lambda)\hat{\text{i}}+(-1-4\lambda)\hat{\text{j}}+(3+2\lambda)\hat{\text{k}}\big]\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)=0$
$(2-2\lambda)(2)+(-1-4\lambda)(4)+(3+2\lambda)(-2)=0$
$4-4\lambda-4-16\lambda-6-4\lambda=0$
$-6-24\lambda=0$
$-24\lambda=6$
$\lambda=-\frac{1}{4}$
Put $\lambda$ in equation (2),
$\vec{\text{a}}=2\lambda\hat{\text{i}}+4\lambda\hat{\text{j}}-2\lambda\hat{\text{k}}$
$=2\Big(-\frac{1}{4}\Big)\hat{\text{i}}+4\Big(-\frac{1}{4}\Big)\hat{\text{j}}-2\Big(-\frac{1}{4}\Big)\hat{\text{k}}$
$\vec{\text{a}}=-\frac{1}2{}\hat{\text{i}}-\hat{\text{j}}+\frac{1}2{}\hat{\text{k}}$
Put the value of $\vec{\text{a}}$ in equation (1),
$\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\Big(-\frac{1}{2}\hat{\text{i}}-\hat{\text{j}}+\frac{1}{2}\hat{\text{k}}\Big)+\vec{\text{b}}$
$\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}+\frac{1}{2}\hat{\text{i}}+\hat{\text{j}}-\frac{1}{2}\hat{\text{k}}$
$=\frac{4\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}+\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}}{2}$
$=\frac{5\hat{\text{i}}+5\hat{\text{k}}}{2}$
$\vec{\text{b}}=\frac{5}{2}\big(\hat{\text{i}}+\hat{\text{k}}\big)$
$\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\Big(-\frac{1}{2}\hat{\text{i}}-\hat{\text{j}}+\frac{1}{2}\hat{\text{k}}\Big)+\frac{5}{2}\big(\hat{\text{i}}+\hat{\text{k}}\big)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "4 Marks" from VECTOR ALGEBRA→VECTOR ALGEBRA — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

A bag contains 4 white and 5 black balls and another bag contains 3 white and 4 black balls. A ball is taken out from the first bag and without seeing its colour is put in the second bag. A ball is taken out from the latter. Find the probability that the ball drawn is white.
→
Solve the following differential equation:
$(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x + 2y}$
→
Find the shortest distance between the lines
$\vec{\text{r}}=6\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}+\lambda\big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=-4\hat{\text{i}}-\hat{\text{k}}+\mu\big(3\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$
→
Show that the lines $\frac{\text{x}-1}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-1}{5}$ and $\frac{\text{x}+2}{4}=\frac{\text{y}-1}{3}=\frac{\text{z}+1}{-2}$ do not intersect.
→
If $\text{A}=\begin{bmatrix}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix},$ show that A-1 = A3.
→
verify that $\text{y}=\text{e}^{\text{m}\cos^{-1}}$ is a solution of the differential equation $(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{m}^2\text{y}=0$
→
Let $\text{f(x)}=\begin{cases}\frac{1-\sin^3\text{x}}{3\cos^2\text{x}},&\text{if }\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if }\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})}^2,&\text{x}>\frac{\pi}{2}\end{cases}$ if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ find a and b.
→
Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisects each other.
→
 find the area of the region bounded by y = |x - 1| and y = 1.
→
Solve the following differential equation:
$\big(\text{y}^2-2\text{xy}\big)\text{dx}=\big(\text{x}^2-2\text{xy}\big)\text{dy}$
→