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Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}$, where $\vec{c}$ is parallel to
$\vec{a}$ and $\vec{d}$ is perpendicular to $\vec{a}$
What is $\vec{c}$ equal to?
Options:
$\vec{a}$ and $\vec{d}$ is perpendicular to $\vec{a}$
What is $\vec{c}$ equal to?
Solution:
1207 Upvotes
Verified Answer
The correct answer is:
$\frac{3(\hat{\mathrm{i}}+\hat{\mathrm{j}})}{2}$
Since $\vec{c}$ is parallel to $\vec{a}$
$\vec{c}=\lambda \vec{a}$
Now $\vec{b}=\vec{c}+\vec{d}=\lambda \vec{a}+\vec{d}$
$=\lambda(\hat{i}+\hat{j})+x \hat{i}+y \hat{j}+z \hat{k}$
$3 \hat{i}+4 \hat{k}=(\lambda+x) \hat{i}+(\lambda+y) \hat{j}+z \hat{k}$
Comparing we get $z=4, \lambda+y=0, \lambda+x=3 \Rightarrow-y+x=3 \quad($ From $(1))$
$\Rightarrow \lambda=-y \quad \ldots(1) \Rightarrow x-y=3$
Now $\vec{d}$ is $\perp \mathrm{r}$ to $\vec{a}$
So, $\cos \theta=0$
$\Rightarrow x+y=0$
Solving $(2)$ and $(3)$ we get
$2 x=3$
$\Rightarrow x=\frac{3}{2}, y=-\frac{3}{2}$
$\Rightarrow \vec{c}=\lambda(\vec{a})=\frac{3}{2}(\hat{i}+\hat{j})$
$\vec{c}=\lambda \vec{a}$
Now $\vec{b}=\vec{c}+\vec{d}=\lambda \vec{a}+\vec{d}$
$=\lambda(\hat{i}+\hat{j})+x \hat{i}+y \hat{j}+z \hat{k}$
$3 \hat{i}+4 \hat{k}=(\lambda+x) \hat{i}+(\lambda+y) \hat{j}+z \hat{k}$
Comparing we get $z=4, \lambda+y=0, \lambda+x=3 \Rightarrow-y+x=3 \quad($ From $(1))$
$\Rightarrow \lambda=-y \quad \ldots(1) \Rightarrow x-y=3$
Now $\vec{d}$ is $\perp \mathrm{r}$ to $\vec{a}$
So, $\cos \theta=0$
$\Rightarrow x+y=0$
Solving $(2)$ and $(3)$ we get
$2 x=3$
$\Rightarrow x=\frac{3}{2}, y=-\frac{3}{2}$
$\Rightarrow \vec{c}=\lambda(\vec{a})=\frac{3}{2}(\hat{i}+\hat{j})$
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