Question

A vector of magnitude 3, bisecting the angle between the vectors = 2i + j – k and

$\overline { \mathrm { b } }$= i – 2j + k and making an obtuse angle with $\overline { \mathrm { b } }$ is –

• A.
• B.
• C.
• D. $\frac { 3 \mathrm { i } - \mathrm { j } } { \sqrt { 10 } }$

Answer 1

Answer: (C)

Answer 2

A vector bisecting the angle between and $\overline { \mathrm { b } }$ is $\frac { \overline { \mathrm { a } } } { | \overline { \mathrm { a } } | } \pm \frac { \overline { \mathrm { b } } } { | \overline { \mathrm { b } } | }$ ; in this case

(i.e.) $\frac { 3 i - j } { \sqrt { 6 } }$ or $\frac { i + 3 j - 2 k } { \sqrt { 6 } }$

A vector of magnitude 3 along these vectors is $\frac { 3 ( 3 \mathrm { i } - \mathrm { j } ) } { \sqrt { 10 } }$ or

Now, $\frac { 3 } { \sqrt { 14 } }$ (i + 3j – 2k). (i – 2j + k) is negative and hence $\frac { 3 } { \sqrt { 14 } }$ (i + 3j – 2k) makes an obtuse angle with $\overline { \mathrm { b } }$ .