1
Question
The area of the triangle having vertices as ^i−2^j+3^k,2^i+3^j−^k,4^i−7^j+7^k is.
The area of the triangle having vertices as ^i−2^j+3^k,2^i+3^j−^k,4^i−7^j+7^k is.
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Solution
The correct option is C 0 sq unit
Let →A=^i−2^j+3^k,
→B=−^i+3^j=^k
and →C=4^i−7^j+7^k
∴→AB=−2^i+3^j−^k−^i+2^j−3^k
=−3^i+5^j−4^k
and →AC=4^i−7^j+7^k−^i+2^j−3^k
=3^i−5^j+4^k
∴ Area of ΔABC=12|→AB×→AC|
=12∣∣
∣
∣∣^i^j^k−35−43−54∣∣
∣
∣∣
=−12∣∣
∣
∣∣^i^j^k3−543−54∣∣
∣
∣∣
=0
[∵ two rows are identical]
Let →A=^i−2^j+3^k,
→B=−^i+3^j=^k
and →C=4^i−7^j+7^k
∴→AB=−2^i+3^j−^k−^i+2^j−3^k
=−3^i+5^j−4^k
and →AC=4^i−7^j+7^k−^i+2^j−3^k
=3^i−5^j+4^k
∴ Area of ΔABC=12|→AB×→AC|
=12∣∣ ∣ ∣∣^i^j^k−35−43−54∣∣ ∣ ∣∣
=−12∣∣ ∣ ∣∣^i^j^k3−543−54∣∣ ∣ ∣∣
=0
[∵ two rows are identical]
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