Show that | a| b + | b| a is perpendicular to | a| b - | b| a , for any two non-zero vectors a and b .

Let c = | a| b + | b| a and d = | a| b - | b| a c d = (| a| b + | b| a) ( ; a| b - | b| a) = | a | 2 b b - | a|| b| b a + | b|| a| a b - | b | 2 a a = | a | 2 | b | 2 - | a|| b| a b + | a|| b| a b - | b | 2 | a | 2 = 0 Hence, c is perpendicular to d .

Vector Algebra — Class 12 Maths Solution

ncert exercise SA NCERT,Page 448,Ex.10.3,Q.No 11

Question

Step-by-step Solution

$= |\vec a{|^2}\vec b \cdot \vec b - |\vec a||\vec b|\vec b \cdot \vec a + |\vec b||\vec a|\vec a \cdot \vec b - |\vec b{|^2}\vec a \cdot \vec a$

$= |\vec a{|^2}|\vec b{|^2} - |\vec a||\vec b|\vec a \cdot \vec b + |\vec a||\vec b|\vec a \cdot \vec b - |\vec b{|^2}|\vec a{|^2} = 0$

Hence, $\vec c$ is perpendicular to $\vec d$.

← All Vector Algebra solutions Practice tests →

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Vector Algebra. Curated by Sachin Sharma. Free for all students.