Show that the vectors a , b , c given by a = i +2 j +3 k , b =2 i + j +3 k and c = i + j + k are non-coplanar. Express vector d =2 i - j -3 k as a linear combination of the vectors a , b and c .

Show that the vectors a , b , c given by a = i +2 j +3 k , b =2 i…

Question

Show that the vectors $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ given by $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ are non-coplanar. Express vector $\vec{\text{d}}=2\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}$ as a linear combination of the vectors $\vec{\text{a}},\ \vec{\text{b}}\text{ and }\vec{\text{c}}$.

✓

Answer

Let the given vectors $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ are coplanar. Then one of the vector is expressible as a linear combination of the other two. Let,
$\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}=\text{x}\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)+\text{y}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
$=\hat{\text{i}}(2\text{x + y})+\hat{\text{j}}(\text{x + y})+\hat{\text{k}}(3\text{x + y})$
$\Rightarrow2\text{x + y}=1,\ \text{x + y}=2,\ 3\text{x + y}=3$
On solving the first two equations we get x = -1, y = 3. Clearly the values of x and y does not satisfy the third equation.
Hence the given vector are non-coplanar.
Now, $\vec{\text{d}}=2\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}$ which can be expressed as
$2\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}=\text{x}\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)\\+\text{y}\big(2\text{i}+\hat{\text{j}}+3\hat{\text{k}}\big)+\text{z}\big(\text{i}+\hat{\text{j}}+\hat{\text{k}}\big)$
$=\text{i}(\text{x}+2\text{y}+\text{z})+\hat{\text{j}}(2\text{x + y + z})+\hat{\text{k}}(3\text{x}+3\text{y + z})$
$\Rightarrow\ \text{x}+2\text{y}+\text{z}=2,\\2\text{x + y + z}=-1,\ 3\text{x}+3\text{y + z}=-3$
$\Rightarrow\ \text{x}=-\frac{8}3,\ \text{y}=\frac{1}3,\ \text{z}=4$
Hence $\vec{\text{d}}$ is expressible as the linear combination of $\vec{\text{a}},\ \vec{\text{b}}\text{ and }\vec{\text{c}}$.