Search any question & find its solution
Question:
Answered & Verified by Expert
If a, be are three non-zero vectors such that each one of them are perpendicular to the sum of the other two vectors, then the value of $\mid \mathbf{a}+\mathbf{b}+\mathbf{c}^{2}$ is
Options:
Solution:
1954 Upvotes
Verified Answer
The correct answer is:
$|a|^{2}+|b|^{2}+\mid q^{2}$
According to the question,
$a \cdot(b+c)=0, b \cdot(\mathfrak{c}+\mathfrak{a})=0, \mathfrak{c} \cdot(\mathbf{a}+\mathbf{b})=0$
$\Rightarrow \quad \quad$ a $\cdot b+a \cdot c=0 \quad \text{...(i)}$
b. $c+b \cdot a=0 \quad \text{...(ii)}$
$c \cdot \mathbf{a}+c \cdot b=0 \quad \text{...(iii)}$
Adding Eqs. (i), (ii) and (iii), we get
$2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} . \mathbf{a})=0 \quad \text{...(iv)}$
Now, $\mid a+b+c^{2}$
$=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+\mid \mathbf{c}^{2}+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})$
$=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+|\mathbb{c}|^{2} \quad$ [using Eq. (iv)]
$a \cdot(b+c)=0, b \cdot(\mathfrak{c}+\mathfrak{a})=0, \mathfrak{c} \cdot(\mathbf{a}+\mathbf{b})=0$
$\Rightarrow \quad \quad$ a $\cdot b+a \cdot c=0 \quad \text{...(i)}$
b. $c+b \cdot a=0 \quad \text{...(ii)}$
$c \cdot \mathbf{a}+c \cdot b=0 \quad \text{...(iii)}$
Adding Eqs. (i), (ii) and (iii), we get
$2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} . \mathbf{a})=0 \quad \text{...(iv)}$
Now, $\mid a+b+c^{2}$
$=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+\mid \mathbf{c}^{2}+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})$
$=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+|\mathbb{c}|^{2} \quad$ [using Eq. (iv)]
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.