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. $\frac { 3 i - j } { \sqrt { 6 } }$
• B.
• C.
• D.

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 $\frac { 2 i + j - k } { \sqrt { 6 } } \pm \frac { i - 2 j + k } { \sqrt { 6 } }$

(i.e.) or

A vector of magnitude 3 along these vectors is or $\frac { 3 ( \mathrm { i } + 3 \mathrm { j } - 2 \mathrm { k } ) } { \sqrt { 14 } }$

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 $\overline { \mathrm { b } }$