Download App

Scalar Or Dot Product — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsScalar Or Dot Product5 Marks
Question
Decompose the vector $6\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}$ into vectors which are parallal and perpendicular to the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}.$
✓

Answer

Let $\vec{\text{a}}=6\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
and $\vec{\text{x}}$ and $\vec{\text{y}}$ be such that
$\vec{\text{a}}=\vec{\text{x}}+\vec{\text{y}}$
$\Rightarrow\vec{\text{y}}=\vec{\text{a}}-\vec{\text{x}}\dots(1)$
Since $\vec{\text{x}}$ is parallel to $\vec{\text{b}},$
$\vec{\text{x}}=\text{t}\vec{\text{b}}$
$\Rightarrow\vec{\text{x}}=\text{t}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=\text{t}\hat{\text{i}}+\text{t}\hat{\text{j}}+\text{t}\hat{\text{k}}\dots(2)$
Substituting the values of $\vec{\text{x}}$ and $\vec{\text{a}}$ in (1), we get
$\vec{\text{y}}=6\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}-\big(\text{t}\hat{\text{i}}+\text{t}\hat{\text{j}}+\text{t}\hat{\text{k}}\big)=(6-\text{t})\hat{\text{i}}+(-3-\text{t})\hat{\text{j}}+(-6-\text{t})\hat{\text{k}}\dots(3)$
Since $\vec{\text{y}}$ is perpendicular to $\vec{\text{b}},$
$\vec{\text{y}}.\vec{\text{b}}=0$
$\Rightarrow\big[(6-\text{t})\hat{\text{i}}+(-3-\text{t})\hat{\text{j}}+(-6-\text{t})\hat{\text{k}}\big].\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=0$
$\Rightarrow1(6-\text{t})+1(-3-\text{t})+1(-6-\text{t})=0$
$\Rightarrow-3-3\text{t}=0$
$\Rightarrow\text{t}=-1$
From (2) and (3), we get
$\vec{\text{x}}=-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{y}}=7\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$
So,
$\vec{\text{a}}=\vec{\text{x}}+\vec{\text{y}}=\big(-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\big(7\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}\big)$

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 "Solve the Following Question.(5 Marks)" from Scalar Or Dot Product→Scalar Or Dot Product — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

Evaluate :

$\int \frac{3 x+4}{\sqrt{2 x^2+2 x+1}} d x$

→
Find the equation of the containing the line $\frac{\text{x}+1}{-3}=\frac{\text{y}-3}{2}=\frac{\text{z}+2}{1}$ and the point $(0, 7, -7)$ and show that the line $\frac{\text{x}}{1}=\frac{\text{y}-7}{-3}=\frac{\text{z}+7}{2}$ also lies in the same plane.
→
If $\text{y}=\text{e}^{\text{x}}\cos\text{x},$ Prvoe that $\frac{\text{dy}}{\text{dx}}=\sqrt{2}\text{e}^\text{x}.\cos\Big(\text{x}+\frac{\pi}{4}\Big)$
→
Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$
→
Solve the following differential equations:
$\text{y}\sqrt{1+\text{x}^2}+\text{x}\sqrt{1+\text{y}^2}\ \frac{\text{dy}}{\text{dx}}=0$
→
Find the area of the region bounded by $\text{y}=\sqrt{\text{x}}$ and $y = x.$
→
Find the equation of the plane passing through the points (-1, 2, 0), (2, 2, -1) and parallel to the line $\frac{\text{x}-1}{1}=\frac{2\text{y}+1}{2}=\frac{\text{z}+1}{-1}.$
→
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{x}^{\frac{2}{3}}+\text{y}^{\frac{2}{3}}=2\text{ at }(1,1)$
→
Evaluate the following integrals:$\int\limits^{\text{a}}_0\frac{1}{\text{x}+\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}$
→
If $\text{x}=\cos\text{t}(3-2\cos^2\text{t}),\text{y}\sin\text{t}(3-2\sin^2\text{t})$ find the value of $\frac{\text{dy}}{\text{dx}}\text{ at t}=\frac{\pi}{4}$
→