Vector Algebra Question 111
Question: If $ a^{2}+b^{2}+c^{2}=1 $ where $ a,b,c\in R $ , then the maximum value of $ {{(4a-3b)}^{2}}+{{(5b-4c)}^{2}}+{{(3c-5a)}^{2}} $ is
Options:
A) 25
B) 50
C) 144
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Let $ {{\vec{r}}_1}=a\hat{i}+b\hat{j}+c\hat{k},{{\vec{r}}_2}=3\hat{i}+4\hat{j}+5\hat{k} $ $ |{{\vec{r}}_1}\times {{\vec{r}}_2}{{|}^{2}}\le |{{\vec{r}}_1}{{|}^{2}}|{{\vec{r}}_2}{{|}^{2}}…(1) $ Now, $ {{\vec{r}}_1}\times {{\vec{r}}_2}= \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ a & b & c \\ 3 & 4 & 5 \\ \end{vmatrix} $ $ =\hat{i}(5b-4c)+\hat{j}(3c-5a)+\hat{k}(4a-3b) $ So, from (1): $ {{(5b-4c)}^{2}}+{{(3c-5a)}^{2}}+{{(4a-3b)}^{2}}\le 50 $
BETA
