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Vector or Cross Product — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsVector or Cross Product1 Mark
MCQ
The vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{k}}$ is to be written as the sum of a vector $\vec{\alpha}$ parallel to $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}$ and a vector $\vec{\beta}$ perpendicular to $\vec{\text{a}}.$ Then $\vec{\alpha}=$
  • ✓
    $\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
  • B
    $\frac{2}{3}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
  • C
    $\frac{1}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
  • D
    $\frac{1}{3}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
✓

Answer

Correct option: A.
$\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
Let:
$\vec{\alpha}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\beta}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$
Now,
$\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{k}}=\vec{\alpha}+\vec{\beta} ($Given$)$
$\Rightarrow3\hat{\text{i}}+0\hat{\text{j}}+4\hat{\text{k}}=(\text{a}_1+\text{b}_1)\hat{\text{i}}+(\text{a}_2+\text{b}_2)\hat{\text{j}}+(\text{a}_3+\text{b}_3)\hat{\text{k}}$
$\Rightarrow\text{a}_1+\text{b}_1=3;\text{a}_2+\text{b}_2=0;\text{a}_3+\text{b}_3=4$
$\Rightarrow\text{a}_1+\text{b}_1=3;\text{a}_2=-\text{b}_2;\text{a}_3+\text{b}_3=4\dots(1)$
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}} ($Given$)$
Also, $\vec{\alpha}$ is parallrl to $\vec{\text{a}}.$
$\Rightarrow\vec{\alpha}\times\vec{\text{a}}=\vec{0}$
$\Rightarrow\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\1&1&0\end{vmatrix}=\vec{0}$
$\Rightarrow-\text{a}_3\hat{\text{i}}+\text{a}_3\hat{\text{j}}+(\text{a}_1-\text{a}_2)\hat{\text{k}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\Rightarrow\text{a}_3=0;\text{a}_1-\text{a}_2=0$
$\Rightarrow\text{a}_3=0;\text{a}_1=\text{a}_2\dots(2)$
Since $\vec{\beta}$ is perpendicular to $\vec{\text{a}},$ we get
$\Rightarrow\vec{\beta}.\vec{\text{a}}=0$
$\Rightarrow\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big).\big(\hat{\text{i}}.\hat{\text{j}}\big)=0$
$\Rightarrow\text{b}_1+\text{b}_2=0$
$\Rightarrow\text{b}_1=-\text{b}_2\dots(3)$
Solving $(1), (2)$ and $(3)$, we get
$\text{a}_1=\frac{3}{2};\text{a}_2=\frac{3}{2};\text{a}_3=0$
$\therefore\vec{\alpha}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$=\frac{3}{2}\hat{\text{i}}+\frac{3}{2}\hat{\text{j}}+0\hat{\text{k}}$
$=\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$

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