Search any question & find its solution
Question:
Answered & Verified by Expert
In a $\triangle A B C,|\mathbf{C B}|=\mathbf{a},|\mathbf{C A}|=\mathbf{b},|\mathbf{A B}|=\mathbf{c}$ and $C D$ is the median through the vertex $C$.
Then, $\mathbf{C A} \cdot \mathbf{C D}=$
Options:
Then, $\mathbf{C A} \cdot \mathbf{C D}=$
Solution:
2799 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{4}\left(a^2+3 b^2-c^2\right)$
As $D$ is mid-point of $A B, A D=\frac{1}{2} \mathbf{A B}$
$\mathbf{C D}=\mathbf{C A}+\mathbf{A D}[\because$ triangle law of vector addition $]$ $\Rightarrow \mathrm{CD}=\mathrm{CA}+\frac{1}{2} \mathbf{A B}$
Now, $\mathbf{C A} \cdot \mathbf{C D}=\mathbf{C A}\left(\mathbf{C A}+\frac{1}{2} \mathbf{A B}\right)$
$\begin{aligned} & =|\mathbf{C A}|^2+\frac{1}{2}(\mathbf{C A} \cdot \mathbf{A B}) \\ & =|\mathbf{C A}|^2+\frac{1}{2}|\mathbf{C A}||\mathbf{A B}| \cos (\pi-A) \\ & =b^2-\frac{1}{2} b c\left(\frac{b^2+c^2-a^2}{2 b c}\right) \\ & =b^2-\left(\frac{b^2+c^2-a^2}{4}\right)=\frac{1}{4}\left(a^2+3 b^2-c^2\right)\end{aligned}$
$\mathbf{C D}=\mathbf{C A}+\mathbf{A D}[\because$ triangle law of vector addition $]$ $\Rightarrow \mathrm{CD}=\mathrm{CA}+\frac{1}{2} \mathbf{A B}$
Now, $\mathbf{C A} \cdot \mathbf{C D}=\mathbf{C A}\left(\mathbf{C A}+\frac{1}{2} \mathbf{A B}\right)$
$\begin{aligned} & =|\mathbf{C A}|^2+\frac{1}{2}(\mathbf{C A} \cdot \mathbf{A B}) \\ & =|\mathbf{C A}|^2+\frac{1}{2}|\mathbf{C A}||\mathbf{A B}| \cos (\pi-A) \\ & =b^2-\frac{1}{2} b c\left(\frac{b^2+c^2-a^2}{2 b c}\right) \\ & =b^2-\left(\frac{b^2+c^2-a^2}{4}\right)=\frac{1}{4}\left(a^2+3 b^2-c^2\right)\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.