# let a=j-k and c=i-j-k. then the vector b satisfying ab+c=0 and a.b=3 is

Let $$\vec{a}=\hat{j}-\hat{k}$$ and$$\vec{c}=\hat{i}-\hat{j}-\hat{k}$$ . Then the vector $$\vec{b}$$ satisfying $$\vec{a} \times \vec{b}+\vec{c}=\overrightarrow{0} \ and \ \vec{a} . \vec{b}=3$$ is :

(1)$$-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$$

(2) $$2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$$

(3)$$\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$$

(4) $$\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$$

from [AIEEE-2010]

## 1 Answer

verified

Option 1 is correct $$-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$$

Explaination::

a x b + c = 0

a x (a x b) + a x c = 0

(a.b)a - (a.a)b + a x c = 0

3a - 2b + (-2i - j - k) = 0

2b = 3(j - k) + (-2i - j - k)

2b=3j-3k-2i-j-k

b=$$-2i+2j-4k\over2$$

b=-i+j-2k

b=$$-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$$